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Calculus Concepts and Modelling: An introduction to the concepts of calculus

Jeremy Sylvestre

Chapter 23 Function limits

Section 23.1 Concept and definition

Subsection 23.1.1 Motivation

Our motivation to create the concept of limit of a sequence was to help investigate trends in summing up more and more terms in a summation pattern that could extend indefinitely, and we will begin investigating such sums in the next chapter. But we can also use sequences to help us investigate trends in functions and understand continuity of functions a little more deeply.
Example 23.1.1. A removable discontinuity.
The function
\begin{equation*} f(t) = \frac{t^3 + 2 t^2 - t - 2}{t - 1} \end{equation*}
is undefined at \(t = 1\text{.}\) But what is the output trend for \(t \approx 1\text{?}\) If we investigate numerically by evaluating \(f(t)\) for inputs that are very close to \(1\text{,}\) we would find that the closer \(t\) is to \(1\text{,}\) the closer \(f(t)\) will be to \(6\text{.}\)
\(t\) \(f(t)\)
\(0.99\) \(5.9501\)
\(0.9999\) \(5.99950001\)
\(0.999999\) \(5.999995000001\)
\(t\) \(f(t)\)
\(1.01\) \(6.0501\)
\(1.0001\) \(6.00050001\)
\(1.000001\) \(6.000005000001\)
The conflicting trends of Example 23.1.2 now put the conclusion of Example 23.1.1 in doubt: how can we be sure that outputs of the function from that example trend toward \(6\text{,}\) if it is possible that different input patterns might exhibit different output trend behaviour. The only way to be sure is to check every possible input pattern. We can do this with sequences of input values.

Subsection 23.1.2 Limit points

To investigate output trends of a function, we need to ensure that the pattern of input values we use are all in the domain of the function. While in the examples in Subsection 23.1.1 involved investigating the behaviour of function outputs for inputs near some \(t\)-value that itself was not in the domain of the function, our investigation would not have been possible if there weren’t at least some domain values \(near\) that \(t\)-value.
Definition 23.1.3. Limit point.
Given a domain set \(D\text{,}\) a number \(c\) is called a limit point of \(D\) when there is at least one example of a sequence \(\nseq{t}\) that satisfies:
  • \(\nseq{t}\) is entirely contained in \(D\)
  • no term of \(\nseq{t}\) is equal to \(c\)
  • \(\displaystyle \nseq{t} \to c\)
The idea here is that \(c\) is a limit point of \(D\) when it is possible to “tiptoe” arbitrarily close to \(c\) from within \(D\) via an example sequence \(\nseq{t} \to c\text{.}\)
Warning 23.1.4.
It is not necessary for the limit point \(c\) itself to be contained in the domain set \(D\text{.}\)
Example 23.1.5. Limit points of intervals.
  1. The number \(c = 2\) is a limit point of both the closed interval \(D_1 = [0,2]\) and the open interval \(D_2 = (0,2)\text{.}\) In both cases, we can use the sequence \(\lrseqtail[n \gt 0]{2 - \frac{1}{n}}\) as our example. This sequence converges to \(2\) but the terms of this sequence are all strictly less than \(2\text{,}\) and so are all contained within both \(D_1\) and \(D_2\text{.}\) Observe that \(2\) is a limit point of \(D_1\) that is itself contained in that domain, but on the other hand \(2\) is a limit point of \(D_2\) that is not itself contained in that domain. Also be careful that in the case of \(D_1\text{,}\) which contains the limit point \(2\text{,}\) the constant sequence \(\nseq{c}\) with \(c_n = 2\) for all \(n\) would not be an appropriate example to use to demonstrate that \(2\) is indeed a limit point, as it does not satisfy the second condition of the definition.
  2. The number \(c = 1\) is a limit point of both the closed interval \(D_1 = [0,2]\) and the open interval \(D_2 = (0,2)\text{.}\) (Can you come up with an example sequence that satisfies the three conditions of the definition in both cases?)
  3. The number \(c = 2.001\) is not a limit point of either \(D_1 = [0,2]\) or \(D_2 = (0,2)\text{.}\) Every sequence with \(\nseq{t} \to 2.001\) would eventually have to contain a tail \(\nseqtail{t}\) that is entirely contained in the open range \((2.0005,2.0015)\text{.}\) But then the terms of that tail would not be contained in either of these domains, and so no sequence converging to \(2.001\) satisfies the requirements to demonstrate that this number is a limit point of either of those domains.

Subsection 23.1.3 Definition and examples

As mentioned before, because function outputs can only be produced by valid input values (in other words, from the domain of the function), we can only investigate a possible trend in output values \(f(t)\) for \(t \approx c\) if it is possible to create a pattern of domain values that trend toward \(c\text{.}\) That is, we require \(c\) to be a limit point of the domain set of function \(f\text{.}\) Then we can consider domain-value sequences \(\nseq{t}\) that converge to \(c\) and investigate the corresponding sequence of output values \(\bbseq{f(t_n)}\text{.}\)
Definition 23.1.6. Limit of a function.
Suppose \(f\) is a function and let \(D\) represent the domain set of \(f\text{.}\) For limit point \(t = c\) of \(D\text{,}\) we say that \(f(t)\) approaches \(L\) as \(t\) approaches \(c\) and write
\begin{equation*} f(t) \to L \qquad \text{as} \qquad t \to c \end{equation*}
when every possible sequence \(\nseq{t}\) that satisifies the three conditions of Limit point for \(c\) (and \(D\)) creates a corresponding output sequence \(\bbseq{f(t_n)}\) that converges to \(L\text{.}\) Alternatively, we may say that \(L\) is the limit of \(f(t)\) as \(t\) approaches \(c\) and write
\begin{equation*} \lim_{t \to c} f(t) = L \text{.} \end{equation*}
If there is no value \(L\) so that the above holds, then we say that the limit of \(f(t)\) as \(t\) approaches \(c\) does not exist.
Remark 23.1.7.
Because a limit point of a set does not itself need to be contained in that set, in the definition above the point \(t = c\) does not need to be in the domain of \(f\) for the notion of a limit of \(f(t)\) as \(t \to c\) to be meaningful. In fact, this is one of the main scenarios where a limit is useful: even if \(f(c)\) is undefined, what can we say about output values \(f(t)\) for \(t \approx c\text{?}\)
And because we only consider input sequences \(\nseq{t}\) for which \(t_n\) is never equal to the limit point \(c\text{,}\) we can make an even stronger remark: a function limit value is independent of the value of the function at the limit point.
Example 23.1.8. Verifying the definition.
Let’s return to Example 23.1.1, where we investigated output trends for \(t \approx 1\) of the function
\begin{equation*} f(t) = \frac{t^3 + 2 t^2 - t - 2}{t - 1} \text{.} \end{equation*}
In that example we only considered two input sequences that converge to \(1\text{:}\)
\begin{equation*} t_n = 1 \pm 10^{-2n-2} \text{.} \end{equation*}
But in both cases we observed that the corresponding output sequence \(\bbseq{f(t_n)}\) appeared to be converging toward \(6\text{.}\) Let’s attempt to confirm that \(f(t) \to 6\) as \(t \to 1\text{.}\)
So suppose \(\nseq{t}\) is an arbitrary sequence with \(\nseq{t} \to 1\) but \(t_n \neq 1\) for all \(n\text{.}\) The corresponding output sequence will have terms
\begin{equation*} f(t_n) = \frac{t_n^3 + 2 t_n^2 - t_n - 2}{t_n - 1} \text{.} \end{equation*}
It is possible to factor the numerator:
\begin{equation*} f(t_n) = \frac{(t_n - 1) (t_n^2 + 3 t_n + 2)}{t_n - 1} \text{.} \end{equation*}
Since we assume \(t_n \neq 1\) we may cancel the common factor in numerator and denominator:
\begin{align*} f(t_n) \amp = \frac{\cancel{(t_n - 1)} (t_n^2 + 3 t_n + 2)}{\cancel{t_n - 1}} \\ \amp = t_n^2 + 3 t_n + 2 \text{.} \end{align*}
Now we may apply the patterns of Pattern 22.3.35 to conclude that since \(\nseq{t} \to 1\text{,}\) we must have
\begin{equation*} \bbseq{f(t_n)} \to 1^2 + 3 \cdot 1 + 2 = 6 \text{.} \end{equation*}
Since our analysis holds for every input sequence that converges to \(1\) (but with no terms equal to \(1\)), the definition of function limit says that
\begin{equation*} \lim_{t \to 1} \frac{t^3 + 2 t^2 - t - 2}{t - 1} = 6 \text{.} \end{equation*}
In practice, to compute a function limit value we are usually less formal than the example above, and we only investigate numerically (as in Example 23.1.1) as a last resort, because it is too easy to be deceived by numerical trends (as we observed in Example 23.1.2).
Example 23.1.9. Practical computation of a function limit.
Our analysis in the previous example really only depended on considering input values close to but not equal to \(1\text{.}\) So if we were not concerned with confirming the limit value rigorously using the formal definition, we could arrive at the actual result
\begin{equation*} \lim_{t \to 1} \frac{t^3 + 2 t^2 - t - 2}{t - 1} = 6 \end{equation*}
via the same algebra but without the input sequence:
\begin{align*} f(t) \amp = \frac{t^3 + 2 t^2 - t - 2}{t - 1} \\ \amp = \frac{\cancel{(t - 1)} (t^2 + 3 t + 2)}{\cancel{t - 1}} \amp \amp\text{(only valid for } t \neq 1\text{)}\\ \amp = t^2 + 3 t + 2 \text{.} \end{align*}
So for \(t \approx 1\) (but \(t \neq 1\)), we have
\begin{align*} f(t) \amp = t^2 + 3 t + 2 \\ \amp \approx 1^2 + 3 \cdot 1 + 2 \\ \amp = 6 \text{.} \end{align*}
From this analysis we can be fairly certain that \(f(t) \to 6\) as \(t \to 1\text{,}\) and if we needed to be completely certain we would then turn to the formal verification presented in Example 23.1.8.
Example 23.1.10. A limit that doesn’t exist.
Consider
\begin{equation*} f(t) = \frac{t + 1}{t - 1} \text{.} \end{equation*}
The domain of \(f\) is all values \(t \neq 1\text{.}\) But we may wish to investigate if there are any trends as \(t \to 1\text{.}\) An input sequence \(\nseq{t}\) in the domain of the function creates an output sequence \(\nseq{s}\) where
\begin{equation*} s_n = f(t_n) = \frac{t_n + 1}{t_n - 1} \text{.} \end{equation*}
We can consider this output sequence as being made up of a numerator sequence \(\seq{t_n + 1}\) and a denominator sequence \(\seq{t_n - 1}\text{.}\) In the case that \(\nseq{t} \to 1\) (but with \(t_n \neq 1\) for all \(n\)), then
\begin{align*} \seq{t_n + 1} \amp \to 2 \text{,} \amp \seq{t_n - 1} \amp \to 0 \text{,} \end{align*}
and so Statement 6 of Pattern 22.3.35 says that \(\nseq{s}\) must diverge. So no matter what potential limit value \(L\) we try, \(\bbseq{f(t_n)} \to L\) is impossible.
Remark 23.1.11.
Effectively, the limit in Example 23.1.10 doesn’t exist because the function exhibits singular behaviour at \(t = 1\text{.}\) We will be more precise about how this type of limit diverges in Subsection 23.5.2.

Section 23.2 Function limit laws

First we establish two fundamental limits for the foundations on which every function is built.
Most of the laws below follow immediately from the facts and laws discussed in Chapter 22.
Repeated application of these laws and the two fundamental limits in Pattern 23.2.1 allow us to handle limits of power functions, polynomial functions, and most rational functions of polynomials.

Warning 23.2.5.

A limit of a quotient function may or may not exist in the case that both the numerator and denominator limits are \(0\text{.}\) For example, in Example 23.1.8 we verified that
\begin{equation*} \lim_{t \to 1} \frac{t^3 + 2 t^2 - t - 2}{t - 1} = 6 \text{,} \end{equation*}
yet separately we have
\begin{align*} \lim_{t \to 1} (t^3 + 2 t^2 - t - 2) \amp = 0 \text{,} \amp \lim_{t \to 1} (t - 1) \amp = 0 \text{.} \end{align*}
As another example, the quotient limit
\begin{equation*} \lim_{t \to 1} \frac{t^3 + 2 t^2 - t - 2}{{(t - 1)}^2} \end{equation*}
also has both numerator and denominator trending toward \(0\text{.}\) But for \(t \neq 1\) we have
\begin{equation*} \frac{t^3 + 2 t^2 - t - 2}{{(t - 1)}^2} = \frac{t^2 + 3 t + 2}{t - 1} \text{,} \end{equation*}
and this simplified ratio has numerator trending toward \(6\) as \(t \to 1\text{,}\) while the denominator trends to \(0\text{,}\) and so the limit as \(t \to 1\) does not exist.
The patterns of the Order Limit Theorem apply to function limits as well.

Justification.

For each of the statements, apply the corresponding statement from Pattern 22.3.37 to the output sequence \(\bbseq{f(t_n)}\) (and the output sequence \(\bbseq{g(t_n)}\) for the third statement), for arbitrary input sequence \(\nseq{t}\) that converges to \(c\) from within \(D\text{.}\)
We can also squeeze function limits.

Justification.

Apply the Pattern 22.3.26 to the output sequences \(\bbseq{f(t_n)}, \bbseq{g(t_n)}, \bbseq{h(t_n)} \) for arbitrary input sequence \(\nseq{t}\) that converges to \(c\) from within \(D\text{.}\)

Section 23.3 One-sided limits

Recall that we have only defined a Limit of a function for \(t \to c\) in the case that \(c\) is a Limit point of the domain of the function. Sometimes we we wish to analyze output trends on a restricted domain.

Example 23.3.1. Limits of a piecewise-defined function.

Consider the function below.
\begin{equation*} f(t) = \begin{cases} 2 t^2 \amp t \lt 1 \\ t - 1 \amp t \gt 1 \end{cases} \end{equation*}
Graph of a discontinuous piecewise function.
The value \(t = 1\) is not in the domain of this function, but it is a limit point of the domain. The sequence \(\nseq{t}\) with
\begin{equation*} t_n = 1 + \frac{{(-1)}^n}{10^n} \end{equation*}
converges to \(1\text{,}\) but the corresponding output sequence \(\bbseq{f(t_n)}\) oscillates between two subsequences:
\begin{equation*} f(t_n) = \begin{cases} {2 \left( 1 - \frac{1}{10^n} \right)}^2 \amp n \text{ odd} \\ \frac{1}{10^n} \amp n \text{ even} \end{cases}\text{.} \end{equation*}
The odd-index subsequence converges to \(2\text{,}\) while the even-index subsequence converges to \(0\text{.}\) Therefore, it’s impossible for the full output sequence \(\bbseq{f(t_n)}\) to converge, as that would violate Pattern 22.3.13. However, the fact that these two subsequences converge provides useful information about output trends of this function for \(t \approx 1\text{.}\)
If we temporarily restrict the domain of our function to \(t \lt 1\text{,}\) then any input sequence \(\nseq{t}\) that converges to \(1\) from within this restricted domain will necessarily satisfy \(t_n \lt 1\) for all \(n\text{.}\) In this case, the corresponding output sequence is \(\seq{2 t_n^2}\text{,}\) which converges to \(2\) if \(\nseq{t} \to 1\text{.}\)
On the other hand, if we temporarily restrict the domain of our function to \(t \gt 1\text{,}\) then any input sequence \(\nseq{t}\) that converges to \(1\) from within this restricted domain will necessarily satisfy \(t_n \gt 1\) for all \(n\text{.}\) In this case, the corresponding output sequence is \(\seq{t_n - 1}\text{,}\) which converges to \(0\) if \(\nseq{t} \to 1\text{.}\)
It would be nice to have notation and terminology to capture the patterns of the previous example in general.

Definition 23.3.2. One-sided limits.

Let \(D\) represent the domain of a function \(f\) and suppose \(c\) is a limit point of \(D\text{.}\) Let \(\leftdomain\) represent only those values in \(D\) that satisfy \(t \lt c\text{,}\) and let \(\rightdomain\) represent only those values in \(D\) that satisfy \(t \gt c\text{.}\)
  1. If \(c\) is a limit point of \(\leftdomain\text{,}\) then we write \(f(t) \to L\) as \(t \to c^{-}\) to mean that \(f(t) \to L\) as \(t \to c\) from within the restricted domain \(\leftdomain\text{.}\) In this case, we say \(f(t)\) approaches \(L\) as \(t\) approaches \(c\) from the left. In limit notation, we write
    \begin{equation*} \lim_{t \to c^{-}} f(t) = L \text{,} \end{equation*}
    and say that \(L\) is the left-hand limit of \(f\) at \(c\).
  2. If \(c\) is a limit point of \(\rightdomain\text{,}\) then we write \(f(t) \to L\) as \(t \to c^{+}\) to mean that \(f(t) \to L\) as \(t \to c\) from within the restricted domain \(\rightdomain\text{.}\) In this case, we say \(f(t)\) approaches \(L\) as \(t\) approaches \(c\) from the right. In limit notation, we write
    \begin{equation*} \lim_{t \to c^+} f(t) = L \text{,} \end{equation*}
    and say that \(L\) is the right-hand limit of \(f\) at \(c\).

Example 23.3.3.

Continuing Example 23.3.1, with our new notation we may write
\begin{align*} \lim_{t \to c^{-}} f(t) \amp = 2 \text{,} \amp \lim_{t \to c^{+}} f(t) \amp = 0 \text{.} \end{align*}

Justification for Statement 1.

(⇒) 
If \(f(t) \to L\) as \(t \to c\text{,}\) then \(f(t_n) \to L\) for every input sequence \(\nseq{t}\) within \(D\) with \(\nseq{t} \to c\text{.}\) In particular, \(f(t_n) \to L\) both when \(\nseq{t} \to c\) from within \(\leftdomain\) and from within \(\rightdomain\text{.}\)
(⇐) 
We assume \(f(t) \to L\) for both \(t \to c^{-}\) and \(t \to c^{+}\text{,}\) and wish to verify that the same is true for \(t \to c\text{.}\) So suppose \(\nseq{t} \to c\) from within \(D\text{.}\) Let \(\nseq{a}\) be the subsequence of those terms \(t_n\) that satisfy \(t_n \lt c\) (reindexed to have indices progress consecutively from \(n = 0\)), and similarly let \(\nseq{b}\) be the subsequence of terms that satisfy \(t_n \gt c\text{.}\) Since we assume that \(t_n \neq c\) for all \(n\text{,}\) the two subsequences together account for all terms of \(\nseq{t}\text{,}\) and so it is also true that the two subsequences \(\bbseq{f(a_n)}\) and \(\bbseq{f(b_n)}\) together account for all terms of the output sequence \(\bbseq{f(t_n)}\text{.}\) Furthermore, since we assume \(f(t) \to L\) as both \(t \to c^{-}\) and \(t \to c^{+}\text{,}\) the two output subsequences \(\bbseq{f(a_n)}\) and \(\bbseq{f(b_n)}\) both converge to \(L\text{.}\) Therefore, we may apply Fact 22.3.20 to conclude that \(\bbseq{f(t_n)} \to L\text{.}\)

Justification for Statement 2.

The definition of \(f(t) \to L\) as \(t \to c\) requires that \(\bbseq{f(t_n)} \to L\) for every input sequence \(\nseq{t} \to c\text{,}\) whether \(\nseq{t} \to c^{+}\) or \(\nseq{t} \to c^{-}\text{.}\)

Example 23.3.5. More limits of a piecewise-defined function.

Here is a slightly modified version of Example 23.3.1.
\begin{equation*} f(t) = \begin{cases} 2 t^2 \amp t \lt 1 \\ t + 1 \amp t \gt 1 \end{cases} \end{equation*}
Graph of a continuous piecewise function.
Consider the “pieces” of \(f\) as separate functions without restricted domains. Applying various Function limit laws, we see that both \(2 t^2 \to 2\) and \(t + 1 \to 2\) as \(t \to 1\text{,}\) and so Fact 23.3.4 tells us that
\begin{align*} \lim_{t \to 1^{-}} 2 t^2 \amp = 2 \text{,} \amp \lim_{t \to 1^+} t + 1 \amp = 2 \text{.} \end{align*}
Now taking the domain restrictions in the defintion of \(f\) into account, the above one-side limits say that
\begin{align*} \lim_{t \to 1^{-}} f(t) \amp = 2 \text{,} \amp \lim_{t \to 1^+} f(t) \amp = 2 \text{.} \end{align*}
Combining these two and applying Fact 23.3.4 again, we have \(f(t) \to 2\) as \(t \to 1\text{.}\) This says that the trend result of output values of \(f(t)\) for \(t \approx 1\) is consistent at \(2\text{,}\) even though those output values will trend toward \(2\) in different ways on each side of input value \(1\) due to the piecewise nature of \(f\text{.}\)
In the example above, it wasn’t really necessary to consider the “pieces” of \(f\) as having unrestricted domains — the Function limit laws are equally valid for one-sided limits.

Section 23.4 An alternative characterization

It is often useful in proofs to have a characterization of the concept of function limit value that does not explicitly refer to sequences.

Justification.

(⇒) 
We would like to prove that if \(f(t) \to L\) as \(t \to c\text{,}\) then values of \(f(t)\) can be restricted to arbitrary open ranges around \(L\) by restricting \(t \approx c\) (but \(t \neq c\)). We will instead prove the contrapositive statement: if values of \(f(t)\) cannot always be restricted to arbitrary open ranges around \(L\text{,}\) then \(f(t) \to L\) as \(t \to c\) is impossible. For this, we will construct an input sequence \(\nseq{t} \to c\) for which the corresponding output sequence \(\bbseq{f(t_n)}\) does not converge to \(L\text{.}\)
So assume that values of \(f(t)\) cannot always be restricted to arbitrary open ranges around \(L\text{.}\) This means that there must be at least one example of an open range \((A,B)\) containing \(L\) that does not contain all output values of \(f(t)\text{,}\) no matter how closely around \(c\) we restrict the input values \(t\text{.}\) For example, the domain \((c - 1, c + 1)\) must contain at least one value \(t_0 \neq c\) for which \(f(t_0)\) is not within \((A,B)\text{.}\) Similarly, the domain \((c - 1/2, c + 1/2)\) must also contain at least one value \(t_1 \neq c\) for which \(f(t_1)\) is not within \((A,B)\text{.}\) Continuing in this fashion, each domain
\begin{equation*} \left( c - \frac{1}{n+1}, c + \frac{1}{n+1} \right) \end{equation*}
must contain at least one value \(t_n \neq c\) for which \(f(t_n)\) is not within \((A,B)\text{.}\)
We leave it to the reader to verify that the sequence \(\nseq{t}\) we have created converges to \(c\text{.}\) But clearly the corresponding output sequence \(\bbseq{f(t_n)}\) cannot converge to \(L\text{,}\) because the open range \((A,B)\) contains \(L\) but contains none of the terms \(f(t_n)\text{.}\) So no tail of \(\bbseq{f(t_n)}\) is contained in that range, and the definition of sequence limit value cannot be confirmed.
Having shown that when values of \(f\) cannot be restricted to every arbitrary open range around \(L\text{,}\) then also \(f(t) \to L\) as \(t \to c\) must be false, we conclude that when \(f(t) \to L\) as \(t \to c\) is actually true, then it must always be possible to restrict values of \(f(t)\) to arbitrary ranges around \(L\) by restricting input values of \(t\) to open subdomains around \(c\text{.}\)
(⇐) 
Assume that values of \(f(t)\) can always be restricted to arbitrary open ranges around \(L\) by restricting input values \(t\) to open subdomains around \(c\) (but excluding \(c\) itself). We would like to verify that \(f(t) \to L\) as \(t \to c\text{.}\) So assume that input sequence \(\nseq{t} \to c\) from within the domain of \(f\) (but, as always, \(t_n \neq c\) for all \(n\)). We need to verify that the corresponding output sequence \(\bbseq{f(t_n)}\) converges to \(L\text{.}\) So consider an open range \((A, B)\) around \(L\text{.}\) From our assumption, it’s possible to specify a corresponding open domain \((a, b)\) around \(c\) so that values of \(f(t)\) are always within \((A,B)\) for \(t\) within \((a,b)\) (but ignoring \(f(c)\)). As \(\nseq{t} \to c\) and \(c\) lies within \((a,b)\text{,}\) there is a tail \(\nseqtail{t}\) that is completely contained in \((a,b)\text{.}\) But within this subdomain all output values of \(f\) lie within range \((A,B)\text{.}\) In particular, the output values \(f(t_n)\) all lie within \((A,B)\) for the terms in the tail is contained in \((a,b)\text{.}\) That is, the corresponding tail \(\bbseqtail{f(t_n)}\) of the output sequence is entirely contained within \((A,B)\text{.}\) Since such a tail of the output sequence can always be determined for every open range \((A,B)\) around \(L\text{,}\) we conclude that \(\bbseq{f(t_n)} \to L\text{.}\)

Section 23.5 Previous concepts as function limits

Many of the concepts we have considered previously can be expressed in terms of function limits.

Subsection 23.5.1 Continuity

Subsubsection Continuity in terms of limits
In Definition 2.4.1, we defined continuity at a point to mean that output values can always be restricted to an open range around a specific output value by restricting the inputs to an open domain around the corresponding specific input value. Through Pattern 23.4.1 we may now connect this to limits.
Fact 23.5.1 effectively says that a function is continuous precisely when computing a limit value coincides with evaluating the function. We can use this to help us compute limit values for compositions of functions involving a continuous function.
Effectively, this pattern says that we may interchange order of computing a limit and evaluating a function:
\begin{equation*} \lim_{t \to c} g \bbrac{f(t)} = g \left( \lim_{t \to c} f(t) \right) \text{.} \end{equation*}
Example 23.5.3.
To compute
\begin{equation*} \lim_{t \to 1} \ln \left( \frac{1 - \sqrt{t}}{1 - t} \right) \text{,} \end{equation*}
we can first compute
\begin{align*} \frac{1 - \sqrt{t}}{1 - t} \amp = \frac{1 - \sqrt{t}}{1 - t} \cdot \frac{1 + \sqrt{t}}{1 + \sqrt{t}} \\ \amp = \frac{\bcancel{1 - t}}{\bcancel{(1 - t)} (1 + \sqrt{t})} \\ \amp = \frac{1}{1 + \sqrt{t}} \\ \amp \to \frac{1}{2} \quad \text{as } t \to 1 \text{.} \end{align*}
Because \(\ln\) is continuous on its domain, we have
\begin{align*} \lim_{t \to 1} \ln \left( \frac{1 - \sqrt{t}}{1 - t} \right) \amp = \ln \left( \lim_{t \to 1} \frac{1 - \sqrt{t}}{1 - t} \right)\\ \amp = \ln \left( \frac{1}{2} \right) \\ \amp = - \ln (2) \text{.} \end{align*}
A power function is continuous on its domain, so we obtain a new limit law by applying Pattern 23.5.2 to limits of powers.
Subsubsection Continuity of combinations of functions
The Function limit laws says that we may reduce limit calculations to limits of the constituent parts of the function involved. Since continuity can be recast in terms of a limit, we can similarly reduce a question of continuity of a function to determining the continuity of the constituent parts of the function.
Justification.
We will only justify Statement 1; the others are similar. So suppose \(f(t)\) and \(g(t)\) are both continuous at \(t = c\text{.}\) Then by Fact 23.5.1 we can say that
\begin{align*} \lim_{t \to c} f(t) \amp = f(c) \text{,} \amp \lim_{t \to c} g(t) \amp = g(c) \text{.} \end{align*}
Applying Statement 2 of Pattern 23.2.2leads to
\begin{align*} \lim_{t \to c} \bigl[ f(t) + g(t) \bigr] \amp = \lim_{t \to c} f(t) + \lim_{t \to c} g(t)\\ \amp = f(c) + g(c) \text{,} \end{align*}
which verifies that the sum function \(f + g\) is continuous at \(t = c\) (again by Fact 23.5.1).
Example 23.5.6. Analyzing continuity in parts.
Consider
\begin{equation*} f(t) = e^t \ln(t) + \sqrt{t} \text{.} \end{equation*}
Individually, all three of the functions
\begin{equation*} e^t, \ln(t), \sqrt{t} \end{equation*}
are continuous at all points \(t \gt 0\text{.}\) So \(e^t \ln(t)\) is a product of continuous functions, and therefore is continuous. Which makes \(f(t)\) a sum of continuous functions, and therefore is also continuous on domain \(t \gt 0\text{.}\)
Subsubsection One-sided continuity
Example 23.5.7.
Consider the function below.
\begin{equation*} f(t) = \begin{cases} t^2 \amp t \le 0 \\ t + 1 \amp t \gt 0 \end{cases} \end{equation*}
Graph of a discontinuous piecewise function.
The function is clearly discontinuous at \(t = 0\text{.}\) However, examining the “flow” of the graph from left to right, the jump discontinuity is not evident until we move past \(t = 0\text{.}\) Indeed, consider \(f\) restricted to the domain \(t \le 0\text{.}\) Let \(g\) represent this restricted-domain version of \(f\text{.}\)
\begin{gather*} g(t) = t^2 \\ t \le 0 \end{gather*}
Graph of one half of a discontinuous piecewise function.
Note that \(c = 0\) is a limit point of the domain
\begin{equation*} \leftdomain = \setdef{t}{t \le 0}\text{.} \end{equation*}
So considering the Limit formulation of continuity, as we clearly have \(g(t) \to g(0) = 0\) as \(t \to 0\) from within domain \(\leftdomain\text{,}\) we may say that \(g\) is continuous at \(t = 0\text{.}\)
Since \(g\) and \(f\) are effectively the same function for \(t \le 0\text{,}\) we should be able to convert this to a statement about the continuity of \(f\) at \(t = 0\text{.}\) But we must qualify this statement to include the domain \(\leftdomain\text{,}\) because \(f\) is discontinuous at \(t = 0\) if we do not include such a qualification. So we will say that \(f\) is continuous from the left at \(t = 0\) because
\begin{equation*} \lim_{t \to 0^{-}} f(t) = f(0)\text{.} \end{equation*}
Definition 23.5.8. One-sided continuity.
Let \(D\) represent the domain of a function \(f\) and suppose \(c\) is a limit point of \(D\text{.}\)
  1. If \(c\) is a limit point of \(\leftdomain\) and \(f(t) \to f(c)\) as \(t \to c^{-}\text{,}\) then we say that \(f\) is continuous at \(t = c\) from the left.
  2. If \(c\) is a limit point of \(\rightdomain\) and \(f(t) \to f(c)\) as \(t \to c^{+}\text{,}\) then we say that \(f\) is continuous at \(t = c\) from the right.
Notation.
When considering continuity on intervals, we will now include/exclude endpoints to indicate one-sided continuity/discontinuity at those endpoints.
Example 23.5.9.
Consider the function below.
\begin{equation*} f(t) = \begin{cases} t^2 \amp t \le -1 \text{,} \\[6pt] {\displaystyle \frac{1}{t^2 - 1}} \amp -1 \lt t \lt 0 \text{,} \\[6pt] {\displaystyle \frac{t-2}{t^2 + t - 6}} \amp t \ge 0 \text{.} \end{cases} \end{equation*}
Graph of a discontinuous piecewise function with continuous pieces.
The parabolic portion of the graph is continuous, including from the left at the domain boundary \(t = -1\text{,}\) so we may say that \(f\) is continuous on the closed interval \(t \le -1\text{.}\) However, the next portion of the graph has a vertical asymptote from the right at \(t = -1\) and has a jump discontinuity at \(t = 0\text{,}\) so we may only say that \(f\) is continuous on the open interval \(-1 \lt t \lt 0\text{.}\) Finally, the graph is continuous from the right at the jump discontinuity at \(t = 0\text{,}\) but there is a removable discontinuity at \(t = 2\text{.}\) So \(f\) is continuous on the half-open interval \(0 \le t \lt 2\) and again on the open interval \(t \gt 2\text{.}\)

Subsection 23.5.2 Long-term and singular behaviour

In Chapter 3, we wrote “\(f(t) \to \infty\) as \(t \to \infty\)” to mean that output values of \(f\) eventually become and stay large and positive. Similar to Fact 23.5.1, we can make this more precise with sequences.
Justification.
We will only consider the case that \(L\) is finite; the arguments in the cases that \(L\) represents one of \(\pm \infty\) are similar.
(⇒) 
Assume that \(f(t) \to L\) as \(t \to \infty\text{,}\) and suppose that \(\nseq{t}\) is a sequence contained in the domain of \(f\) so that \(\nseq{t} \to \infty\text{.}\) We would like to verify that \(\bbseq{f(t_n)} \to L\text{,}\) so suppose \((a,b)\) is an open range around \(L\text{.}\) Our assumption about the long-term behaviour of \(f\) means that values of \(f(t)\) eventually become and stay arbitrarily close to \(L\text{,}\) so there is some open domain \((M,\infty)\) where all values of \(f(t)\) land within \((A,B)\) for \(t\) in \((M,\infty)\text{.}\) But as \(\nseq{t} \to \infty\text{,}\) there is a tail \(\nseqtail{t}\) entirely contained within \((M,\infty)\text{,}\) and therefore the corresponding tail \(\bbseqtail{f(t_n)}\) is entirely contained within \((a,b)\text{.}\)
(⇐) 
Similarly to the justification of Pattern 23.4.1, one can instead demonstrate the contrapositive: if \(f(t) \not\to L\) as \(t \to \infty\text{,}\) then it is not true that output sequences \(\bbseq{f(t_n)}\) will always converge to \(L\) for input sequences that diverge to infinity. We leave the details of constructing such an example input sequence \(\nseq{t}\) to the reader.
Remark 23.5.11.
Fact 23.5.10 is similarly true about ancient behaviour by considering \(t \to -\infty\) versus sequences \(\nseq{t}\) that diverge to negative infinity.
Singular behaviour is simply an inversion of long-term behaviour, and we can similarly take a limit point of view by considering output sequences diverging to infinity instead of input sequences diverging to infinity. The justification for the following fact is similar to arguments we’ve already made in this chapter, and will be omitted this time.
Remark 23.5.13.
Similarly to Fact 23.5.12, we may write
\begin{align*} \lim_{t \to c^{+}} f(t) \amp = \pm \infty \amp \lim_{t \to c^{-}} f(t) \amp = \pm \infty \end{align*}
by only considering sequences contained in the domain \(t \gt c\) (for the limit \(t \to c^{+}\)) or only considering sequences contained in the domain \(t \lt c\) (for the limit \(t \to c^{-}\)).

Subsection 23.5.3 Definite integrals

Recall that a definite integral value
\begin{equation*} \ccmint{a}{b}{f(t)}{t} \end{equation*}
represents the unique value (if it exists) that sits between every pair of upper and lower sum estimates:
\begin{equation*} L(P) \le \ccmint{a}{b}{f(t)}{t} \le U(P) \end{equation*}
for every partition \(P\) of the integration domain \(a \le t \le b\text{.}\) If \(f\) is already known to be integrable (for example, if \(f\) is continuous on the integration domain), then it is sufficient to only consider regular partition (that is, creating a regular step function approximation to \(f(t)\) on that domain). Since the first “step” in creating a regular step function is to decide how many steps to use, in effect we create two sequences Riemann sum values indexed by \(n\text{,}\) the number of steps to use:
  • a sequence \(\nseq{L}\) where each term \(L_n\) represents the lower sum approximation value using a step function with \(n\) steps
  • a sequence \(\nseq{U}\) where each term \(U_n\) represents the upper sum approximation value using a step function with \(n\) steps.
Example 23.5.14.
For
\begin{equation*} \ccmint{0}{1}{t^2 - 2 t + 2}{t} \text{,} \end{equation*}
the integrand function \(f(t) = t^2 - 2 t + 2\) is decreasing on the domain of integration, and so in this case lower sums will be right Riemann sums and upper sums will be left Riemann sums.
A graph depicting a right Riemann sum for a decreasing function creating a lower sum.
(a) Lower sum.
A graph depicting a left Riemann sum for a decreasing function creating an upper sum.
(b) Upper sum.
Figure 23.5.15.
Following Procedure 6.1.6 twice, choosing sample points to be right endpoints in the lower sum case and left endpoints in the upper sum case, we obtain
\begin{align*} L_n \amp = \sum_{k = 1}^{n} f\left(\frac{k}{n}\right) \cdot \frac{1}{n} \amp U_n \amp = \sum_{k = 1}^{n} f\left(\frac{k - 1}{n}\right) \cdot \frac{1}{n} \\ \amp = \sum_{k = 1}^{n} \left( \frac{k^2}{n^2} - \frac{2 k}{n} + 2 \right) \cdot \frac{1}{n} \text{,} \amp \amp = \sum_{k = 1}^{n} \left(\frac{(k - 1)^2}{n^2} - \frac{2 (k - 1)}{n} + 2\right) \cdot \frac{1}{n} \text{.} \end{align*}
If we used the patterns of Subsection 5.3.2 to simplify these expressions, eventually we would be faced with having to calculate sums like
\begin{align*} \amp \sum_{k = 1}^{n} k^2 \amp \amp \sum_{k = 1}^{n} k \end{align*}
for which there are known formulas in \(n\text{.}\) Let’s get Sage to apply those patterns and formulas for us.
According to Sage, we have
\begin{align*} L_n \amp = \frac{8 n^2 - 3 n + 1}{6 n^2} \text{,} \amp U_n \amp = \frac{8 n^2 + 3 n + 1}{6 n^2} \text{.} \end{align*}
With the only difference being a plus/minus sign switch between them, clearly the lower sum result is always less than the corresponding upper sum result, as expected, and the actual definite integral value will lie between these two values for every value of \(n\text{.}\)
We know that upper and lower sum values “squeeze” the exact definite integral value (Fact 6.2.1). We may now take a sequence perspective on this idea.
Notationally we can take the idea of these sequences “squeezing” down to the value of the definite integral further: if we let \(A\) represent the value of
\begin{equation*} \ccmint{a}{b}{f(t)}{t} \text{,} \end{equation*}
then we might write
\begin{align*} \nseq{L} \amp \to A^{-} \text{,} \amp \nseq{U} \amp \to A^{+} \text{.} \end{align*}
Example 23.5.17.
Continuing Example 23.5.14, for \(f(t) = t^2 - 2 t + 2\) on domain \(0 \le t \le 1\) we had
\begin{align*} L_n \amp = \frac{8 n^2 - 3 n + 1}{6 n^2} \text{,} \amp U_n \amp = \frac{8 n^2 + 3 n + 1}{6 n^2} \text{.} \end{align*}
Using our strategy of “dominant terms”, we see that both of these sequences converge to \(8 / 6\) (or \(4 / 3\)), which agrees with the result for the definite integral that we would obtain from Pattern 7.4.2:
\begin{align*} \ccmint{0}{1}{t^2 - 2 t + 2}{t} \amp = \funceval{\frac{t^3}{3} - 2 \cdot \frac{t^2}{2} + 2 t}{t = 1} \\ \amp = \frac{1}{3} - 2 \cdot \frac{1}{2} + 2 \\ \amp = \frac{4}{3}\text{.} \end{align*}
Example 23.5.18.
Analyzing
\begin{equation*} \ccmint{0}{1}{t^2 - t + 1} \end{equation*}
via upper and lower sums seems like it should be just as simple as our previous example, but in this case our integrand function \(f(t) = t^2 - t + 1\) is sometimes decreasing and sometimes increasing, which means the sample points will not remain consistently at one endpoint or another, but instead will flip.
We could exploit the symmetry of the graph to simplify things, but instead let’s exercise our reasoning skills and exploit the symmetry of the positioning within the domain of the critical point at \(t = 1/2\text{,}\) where the graph switches from decreasing to increasing. How the endpoints of our regular partition fall relative to this critical point depends on whether the number of steps \(n\) is even or odd, and so we will break our analysis up into those two cases.
Even number of steps.
In this case, the critical point at \(t = 1/2\) will always line up with a partition boundary, so we may split the domain in two: for the first half of the domain, lower sum sample points will be right endpoints and upper sum sample points will be left endpoints, and for the second half of the domain this will reverse.
A graph depicting a lower sum for an even number of partition cells.
(a) Lower sum for even \(n\text{.}\)
A graph depicting an upper sum for an even number of partition cells.
(b) Upper sum for even \(n\text{.}\)
Figure 23.5.19.
So when \(n\) is even we have
\begin{align*} L_n \amp = \sum_{k = 1}^{n/2} f\left(\frac{k}{n}\right) \cdot \frac{1}{n} + \sum_{k = n/2 + 1}^{n} f\left(\frac{k - 1}{n}\right) \cdot \frac{1}{n} \text{,}\\ U_n \amp = \sum_{k = 1}^{n/2} f\left(\frac{k - 1}{n}\right) \cdot \frac{1}{n} + \sum_{k = n/2 + 1}^{n} f\left(\frac{k}{n}\right) \cdot \frac{1}{n} \text{.} \end{align*}
Odd number of steps.
In this case, the critical point at \(t = 1/2\) will always line up with a midpoint of a partition subdomain, and so this time we will split the domain in three: for the first “half” of the domain (excluding the “middle” parition subdomain), the second “half” of the domain, and the “middle” partition subdomain by itself. On the two “halves”, upper/lower sum sample points line up with left/right endpoints exactly as in the previous case. But on that middle partition subdomain that straddles the critical point, the lowest value occurs at \(t = 1/2\text{,}\) while the highest value occurs symmetrically at the two endpoints.
A graph depicting a lower sum for an odd number of partition cells.
(a) Lower sum for odd \(n\text{.}\)
A graph depicting an upper sum for an odd number of partition cells.
(b) Upper sum for odd \(n\text{.}\)
Figure 23.5.20.
So when \(n\) is odd we have
\begin{align*} L_n \amp = \sum_{k = 1}^{(n-1)/2} f\left(\frac{k}{n}\right) \cdot \frac{1}{n} + f\left(\frac{1}{2}\right) \cdot \frac{1}{n} + \sum_{k = (n+3)/2}^{n} f\left(\frac{k - 1}{n}\right) \cdot \frac{1}{n} \text{,}\\ U_n \amp = \sum_{k = 1}^{(n-1)/2} f\left(\frac{k - 1}{n}\right) \cdot \frac{1}{n} + \sum_{k = (n+1)/2}^{n} f\left(\frac{k}{n}\right) \cdot \frac{1}{n} \text{.} \end{align*}
Note that the “middle” partition subdomain is indexed by \(k = (n+1)/2\text{.}\) For the upper sum we have chosen the right endpoint as the sample point for that subdomain, and so have absorbed that term into the sum for the second “half” of partition cells.
Effectively we are analyzing subsequences of the upper and lower sum sequences. Since in each case the even/odd-index subsequences cover all terms, if these subsequences all converge to the same value, we can be confident that the sequences themselves converge to the same value.
Convergence of lower sum even subsequence.
When \(n\) is even, we may assume that it is double another integer: \(n = 2 m\text{.}\) Substituting this into our previous expression for \(L_n\) in the even case, we have
\begin{equation*} L_n = \sum_{k = 1}^{m} f\left(\frac{k}{2 m}\right) \cdot \frac{1}{2 m} + \sum_{k = m + 1}^{2 m} f\left(\frac{k - 1}{2 m}\right) \cdot \frac{1}{2 m}\text{.} \end{equation*}
Let’s put this into Sage to simplify.
Analyzing dominant terms of the result
\begin{equation*} L_m = \frac{20 m^2 - 3 m + 1}{24 m^2} \end{equation*}
we conclude \(\seq{L_m} \to 5 / 6\text{,}\) which agrees with the result of the definite integral that we could compute by other “shortcut” methods.
We leave the analysis of the other three cases to you, the reader.

Subsection 23.5.4 The derivative

Reviewing the definition of derivative value Definition 11.5.1, it attempts to describe the pattern of difference quotient values at a specific point \(t = t^\ast\) for \(\change{t} \approx 0\text{.}\) So it is not a limit of the function \(q\) whose derivative values we are computing, but instead a limit of a new function whose independent variable is \(\change{t}\text{:}\)
\begin{gather} Q_{t^\ast}(\change{t}) = \frac{\change{q}}{\change{t}} = \frac{q(t^\ast + \change{t}) - q(t^\ast)}{\change{t}}\text{,}\tag{✶} \end{gather}
where \(t^\ast\) is a specific point in the domain of \(q\text{.}\)
Justification.
Re-reading the definition of Definition 11.5.1, interpret “arbitrarily close to \(d\)” to mean “restricted within some small open range around \(d\)”, and interpret “taking \(\change{t}\) sufficiently small” to mean “restricted within some small open domain around \(0\) (but \(\change{t} \neq 0\)).” Then the statement at hand is equivalent to Pattern 23.4.1 applied to \(Q_{t^\ast}(\change{t})\text{.}\)
A difference quotient function \(Q_{t^\ast}(\change{t})\) always “evaluates” to an indeterminate \(0/0\) form, similar to many of the function limit examples we’ve considered in this chapter. In particular, you may have noticed a similarity between Example 23.1.9 and our algebraic approach to computing derivative values from the definition (Procedure 12.2.2). Fact 23.5.21 says that this is not a coincidence.
Example 23.5.22. Computing a derivative value using a limit.
Let’s repeat Example 12.2.1, but only for the specific derivative value at \(t = 2\text{.}\) To maintain consistency with the notation used in this subsection, we’ll write \(q(t)\) for our position function instead of \(s(t)\text{:}\)
\begin{equation*} q(t) = -2 t^2 + 6 t \text{.} \end{equation*}
We would like to compute the value of \(q'(2)\) using a limit. So form the associated difference quotient function
\begin{align*} Q_2(\change{t}) \amp = \frac{q(2 + \change{t}) - q(2)}{\change{t}} \\ \amp = \frac{\bbrac{-2 {(2 + \change{t})}^2 + 6 (2 + \change{t})} - (-2 \cdot 2^2 + 6 \cdot 2)}{\change{t}} \end{align*}
and compute its limit as \(\change{t} \to 0\) through algebraic simplification of the difference quotient function \(Q_2(\change{t})\text{:}\)
\begin{align*} q'(2) \amp = \lim_{\change{t} \to 0} Q_2(\change{t}) \\ \amp = \lim_{\change{t} \to 0} \frac{\bbrac{-2 {(2 + \change{t})}^2 + 6 (2 + \change{t})} - (-2 \cdot 2^2 + 6 \cdot 2)}{\change{t}} \\ \amp = \lim_{\change{t} \to 0} \frac{-2 \cancel{\change{t}} + {(\change{t})}^{\cancel{2}}}{\cancel{\change{t}}} \\ \amp = \lim_{\change{t} \to 0} -2 + \change{t} \\ \amp = -2 \text{.} \end{align*}
You may verify that this agrees with the value of \(q'(2)\) that you might compute using other “shortcut” methods.

Section 23.6 L’Hopital’s rule

Subsection 23.6.1 \(0/0\) forms

The Function limit laws say that a limit of a quotient
\begin{equation*} \lim_{t \to c} \frac{f(t)}{g(t)} \end{equation*}
is \(0\) if the numerator function approaches \(0\) but the denominator function doesn’t, and doesn’t exist if the denominator function approaches \(0\) but the numerator function doesn’t. What if both \(f(t) \to 0\) and \(g(t) \to 0\) as \(t \to c\text{?}\)
Definition 23.6.1. Indeterminate form \(0/0\).
A limit of the form
\begin{equation*} \lim_{t \to c} \frac{f(t)}{g(t)} \end{equation*}
where both \(f(t) \to 0\) and \(g(t) \to 0\) as \(t \to c\text{.}\)
Example 23.6.2.
  1. Consider
    \begin{align*} f_1(t) \amp = t^2 - 1 \text{,} \amp g_1(t) \amp = {(t - 1)}^2 \text{.} \end{align*}
    Both functions approach \(0\) as \(t \to 1\text{,}\) but
    \begin{equation*} \lim_{t \to 1} \frac{f_1(t)}{g_1(t)} \end{equation*}
    does not exist.
  2. Consider
    \begin{align*} f_2(t) \amp = {(t - 1)}^2 \text{,} \amp g_2(t) \amp = t^2 - 1 \text{.} \end{align*}
    Again both functions approach \(0\) as \(t \to 1\text{,}\) but
    \begin{equation*} \lim_{t \to 1} \frac{f_2(t)}{g_2(t)} \end{equation*}
    exists and is equal to \(0\text{.}\)
  3. Consider
    \begin{align*} f_3(t) \amp = t^2 - 1 \text{,} \amp g_3(t) \amp = t - 1 \text{.} \end{align*}
    Both functions approach \(0\) as \(t \to 1\text{,}\) but
    \begin{equation*} \lim_{t \to 1} \frac{f_3(t)}{g_3(t)} \end{equation*}
    exists and is equal to \(2\text{.}\)
Example 23.6.2 demonstrates that a limit of indeterminate form \(0/0\) may have any result. Though we already knew this, since every derivative calculation involves a \(0/0\) indeterminate form limit, and derivative values frequently exist and can take on many different values. In fact, known derivative formulas give us an alternative way to compute \(0/0\)-form limits.
Example 23.6.3.
Consider
\begin{equation*} \lim_{t \to 0} \frac{e^t - 1}{t} \text{.} \end{equation*}
Since the numerator is a continuous function, we have \(e^t - 1 \to e^0 - 1 = 0\text{.}\) So the limit is a \(0/0\) indeterminate form. But using
Using Fact 23.5.21, we have
\begin{align*} \funceval{\ddt{(e^t)}}{t = 0} \amp = \lim_{\change{t} \to 0} \frac{e^{(0 + \change{t})} - e^0}{\change{t}}\\ \amp = \lim_{\change{t} \to 0} \frac{e^{\change{t}} - 1}{\change{t}} \text{.} \end{align*}
Ignoring the difference in variables, this is exactly the limit we would like to compute. Since
\begin{equation*} \ddt{(e^t)} = e^t \text{,} \end{equation*}
we have
\begin{align*} \lim_{t \to 0} \frac{e^t - 1}{t} \amp = \lim_{\change{t} \to 0} \frac{e^{\change{t}} - 1}{\change{t}}\\ \amp = \funceval{\ddt{(e^t)}}{t = 0} \\ \amp = \funceval{e^t}{t = 0} \\ \amp = 1 \text{.} \end{align*}
Can derivatives help us compute other \(0/0\)-form limits? Remember that a function limit measures how output-value sequences behave for all possible input-value sequences that converge to the limit point. So, similarly to Section 3.2, in some sense a \(0/0\)-form limit is like a “race” to zero between the output-value sequences of the numerator function and the denominator function. In the first limit of Example 23.6.2, the denominator function was approaching zero faster and “won” the race, causing the limit of their ratio to diverge to infinity (though differently-signed infinities from different sides of the limit point). In the second limit of Example 23.6.2, the roles of numerator and denominator were reversed and the numerator function “won” the race, causing the limit of their ratio to converge to \(0\text{.}\) And in the third limit of Example 23.6.2, both numerator and denominator functions were approaching zero at the same rate, and so these zeros “cancelled out” before they actually reached zero, leaving behind a constant ratio of \(2\text{.}\)
A derivative measures a rate of variation, so we can use the derivatives of the numerator and denominator functions to compare how quickly they are trending toward zero near the limit point.
Warning 23.6.5.
  • Do not apply L’Hôpital’s Rule to limits to which it does not apply! The limit must be a \(0/0\)-form (or another indeterminate form — see Subsection 23.6.2) for L’Hôpital’s Rule to apply.
  • If you have learned something called the “quotient rule” for computing derivatives in a previous calculus course, L’Hôpital’s Rule does not involve the “quotient rule.” The expression in L’Hôpital’s Rule is a quotient of derivatives, not a derivative of a quotient.
Notation.
We will use a “l’H” indicator over an equals sign between two limits when L’Hôpital’s Rule has been applied:
\begin{equation*} \lim_{t \to c} \frac{f(t)}{g(t)} \lhopeq \lim_{t \to c} \frac{f'(t)}{g'(t)} \text{.} \end{equation*}
Example 23.6.6.
In each of these, where L’Hôpital’s Rule has been applied you should verify that the preceding limit was a \(0/0\)-form.
  1. We have
    \begin{equation*} \lim_{t \to 1} \frac{t^2 - 1}{{(t - 1)}^2} \lhopeq \lim_{t \to 1} \frac{2 t}{2 (t - 1)} \end{equation*}
    but this second limit does not exist because the denominator trends to \(0\) while the numerator does not.
  2. \(\displaystyle \displaystyle \lim_{t \to 1} \frac{{(t - 1)}^2}{t^2 - 1} \lhopeq \lim_{t \to 1} \frac{\cancel{2} (t - 1)}{\cancel{2} t} = \frac{1 - 1}{1} = 0\)
  3. \(\displaystyle \displaystyle \lim_{t \to 1} \frac{t^2 - 1}{t - 1} \lhopeq \lim_{t \to 1} \frac{2 t}{1} = \frac{2 \cdot 1}{1} = 2\)
  4. \(\displaystyle \displaystyle \lim_{t \to 0} \frac{\sin(t)}{t} \lhopeq \lim_{t \to 0} \frac{\cos(t)}{1} = \frac{\cos(0)}{1} = 1\)
Remark 23.6.7.
If the new limit you obtain after applying L’Hôpital’s Rule is still \(0/0\) form, you may apply the rule again. But if you apply it many times and still get \(0/0\) form every time, you might look for any patterns that might tell you whether you are actually making progress toward a final result.
Example 23.6.8. Applying L’Hôpital’s Rule multiple times.
In this example, the decreasing degree in the denominator as we repeatedly apply L’Hôpital’s Rule indicates that we will eventually come to a determinate form. Once again, where \(\lhopeq\) appears you should verify that the preceding limit was \(0/0\)-form.
\begin{align*} \lim_{t \to 0} \frac{\tan(t) - t}{t^3} \amp \lhopeq \lim_{t \to 0} \frac{\sec^2(t) - 1}{3 t^2}\\ \amp \lhopeq \lim_{t \to 0} \frac{2 \sec(t) \cdot \bbrac{\sec(t) \tan(t)} - 0}{6 t} \\ \amp = \lim_{t \to 0} \frac{2 \sec^2(t) \tan(t)}{6 t} \\ \amp \lhopeq \lim_{t \to 0} \frac {\bbrac{2 \sec^2(t)} \cdot \sec^2(t) + \Bbrac{4 \sec{t} \cdot \bbrac{\sec(t) \tan(t)}} \cdot \tan(x)} {6}\\ \amp = \frac {\bbrac{2 {(1)}^2} \cdot {(1)}^2 + \bbrac{4 \cdot 1 \cdot (1 \cdot 0)} \cdot 0} {6}\\ \amp = \frac{1}{3} \end{align*}

Subsection 23.6.2 Other indeterminate forms

Recall that \(f(t) \to 0\) as \(t \to c\) means that for every input sequence \(\nseq{t}\) that converges to \(c\text{,}\) the corresponding output sequence \(\bbseq{f(t_n)}\) converges to \(0\text{.}\) So when we encounter a \(0/0\)-form limit, it is (usually) not actually \(0/0\text{,}\) it is a trend of \(f(t_n)/g(t_n)\) values with both \(f(t_n) \approx 0\) and \(g(t_n) \approx 0\text{.}\) So instead of \(0/0\text{,}\) it is better to think of this indeterminate form as “small over small”. And the reason this form is “indeterminate” is that we cannot reliably determine the result of “small over small”:
  • if the numerator becomes small faster than the denominator, then the ratio will become small
  • if the denominator becomes small faster than the numerator, then the ratio will become large
  • if the numerator and denominator become small at the same rate, then the ratio will be approximately constant.
Here are some other indeterminate limit forms.
Subsubsection \(\infty/\infty\) form
Definition 23.6.9. Indeterminate form \(\infty/\infty\).
A limit of the form
\begin{equation*} \lim_{t \to c} \frac{f(t)}{g(t)} \end{equation*}
where both \(f(t) \to \infty\) and \(g(t) \to \infty\) as \(t \to c\text{.}\)
Note 23.6.10.
We will also refer to a limit as \(\infty/\infty\) form if either or both of \(f\) and \(g\) trend to \(-\infty\text{.}\)
Similarly to the above discussion, we cannot reliably determine the result of “large over large” because there are multiple possible outcomes:
  • large ratio result if the numerator becomes large faster than the denominator
  • small ratio result if the denominator becomes large faster than the numerator
  • approximately constant ratio result if the numerator and denominator become large at the same rate.
We could apply L’Hôpital’s Rule for \(0/0\) form by first performing an algebraic manipulation: if
\begin{equation*} \lim_{t \to c} \frac{f(t)}{g(t)} \end{equation*}
is \(\infty/\infty\) form, then
\begin{equation*} \lim_{t \to c} \frac{1/g(t)}{1/f(t)} \end{equation*}
is \(0/0\) form, and the two limits are equivalent. Luckily, however, L’Hôpital’s Rule can be applied directly to an \(\infty/\infty\)-form limit.
Example 23.6.12.
Both \(e^t \to \infty\) and \(t^2 \to \infty\) as \(t \to \infty\text{,}\) so we may apply the \(\infty/\infty\)-form version of L’Hôpital’s Rule:
\begin{align*} \lim_{t \to \infty} \frac{e^t}{t^2} \amp \lhopeq \lim_{t \to \infty} \frac{e^t}{2 t}\\ \amp \lhopeq \lim_{t \to \infty} \frac{e^t}{2} \\ \amp = \infty \text{.} \end{align*}
Warning 23.6.13.
We will repeat a previous warning: do not apply L’Hôpital’s Rule to limits to which it does not apply! For example, it would be incorrect to apply L’Hôpital’s Rule to the limit in Example 23.6.12 if it were for \(t \to -infty\) instead of \(t \to \infty\text{.}\)
Subsubsection \(0 \cdot \infty\) form
Again this form of limit is indeterminate because we cannot reliably determine the result of “small times large”. (You are encouraged to run through the different possible outcomes on your own.)
Example 23.6.15.
The limit
\begin{equation*} \lim_{t \to 0^{+}} \sqrt{t} \ln(t) \end{equation*}
is \(0 \cdot (\infty)\) form. So instead consider the equivalent limits
\begin{align*} \amp \lim_{t \to 0^{+}} \frac{\sqrt{t}}{1/\ln(t)} \amp \amp \lim_{t \to 0^{+}} \frac{\ln(t)}{1/\sqrt{t}}\text{.} \end{align*}
The first limit is now \(0/0\) form and the second is \(-\infty/\infty\) form. If we applied L’Hôpital’s Rule to the first limit, the new limit would still have a logarithm in it because the derivative of \(1/\ln(t)\) would require a Chain Rule. However, the second limit will no longer have a logarithm after applying L’Hôpital’s Rule:
\begin{align*} \lim_{t \to 0^{+}} \sqrt{t} \ln(t) \amp = \lim_{t \to 0^{+}} \frac{\ln(t)}{1/\sqrt{t}}\\ \amp = \lim_{t \to 0^{+}} \frac{\ln(t)}{t^{-1/2}} \\ \amp \lhopeq \lim_{t \to 0^{+}} \frac{1/t}{(-1/2) t^{-3/2}} \\ \amp = \lim_{t \to 0^{+}} -2 t^{1/2} \\ \amp = 0 \text{.} \end{align*}
Subsubsection \(\infty - \infty\) form
This is an indeterminate form because we cannot reliably determine the result of “large minus large”. (In this case, it is important that both “large” terms are “large and positive.”) Consider the possibilities:
  • large, positive difference result if the first term becomes large faster than the second
  • large, negative difference result if the second term becomes large faster than the first
  • approximately constant difference result if the two terms large at the same rate.
Example 23.6.17.
The limit
\begin{equation*} \lim_{t \to {\frac{\pi}{2}}^{-}} \tan(t) - \sec(t) \end{equation*}
is an \((\infty - \infty)\)-form limit. We should not apply L’Hôpital’s Rule directly to this limit, because that rule does not apply. However, if we convert these trig functions into sines and cosines, we can manipulate the expression into a \(0/0\)-form limit to which L’Hôpital’s Rule can be applied:
\begin{align*} \lim_{t \to {\frac{\pi}{2}}^{-}} \tan(t) - \sec(t) \amp = \lim_{t \to {\frac{\pi}{2}}^{-}} \frac{\sin(t)}{\cos{t}} - \frac{1}{\cos(t)}\\ \amp = \lim_{t \to {\frac{\pi}{2}}^{-}} \frac{\sin(t) - 1}{\cos{t}} \\ \amp \lhopeq \lim_{t \to {\frac{\pi}{2}}^{-}} \frac{\cos(t) - 0}{-\sin{t}} \\ \amp = \frac{0}{-1} \\ \amp = 0 \text{.} \end{align*}
Subsubsection Indeterminate forms involving exponents
Limits of the form
\begin{align*} \amp 0^0 \amp \amp \infty^0 \amp \amp 1^\infty \end{align*}
are also indeterminate. (Remember: the “values” in these forms do not represent contant values, they represent results of output sequence trends.) For example, the output sequence values for the expression in a \(1^\infty\)-form limit could be:
  • trending to \(1\) because the base values trend to \(1\text{,}\) and \(1^a = 1\) for all values of \(a\)
  • becoming large if the base values trend to \(1\) from above, because the exponent is trending to large values, and a base greater than \(1\) raised to large exponent will create a large result
  • becoming small if the base values trend to \(1\) from below, because the exponent is trending to large values, and a base less than \(1\) raised to large exponent will create a small result.
As we have done in the past, we may exploit the algebraic properties of logarithms to help us deal with exponents.
Justification.
Because the exponential function is continuous, Pattern 23.5.2 tells us that
\begin{equation*} \lim_{t \to c} g(t) = L \qquad \implies \qquad \lim_{t \to c} \exp\bbrac{g(t)} = \exp(L) \text{.} \end{equation*}
Apply this pattern with \(g(t)\) replaced by \(\ln\bbrac{f(t)}\text{,}\) noting that
\begin{equation*} \exp\Bbrac{\ln\bbrac{f(t)}} = f(t) \text{.} \end{equation*}
Example 23.6.19.
It seems like we should have
\begin{equation*} \lim_{t \to 1} t^t = 1 \text{,} \end{equation*}
but can we be sure? We don’t have any limit laws that directly address this situation. (Note that Pattern 23.5.4 does not apply here.)
Instead, let’s try to compute
\begin{equation*} \lim_{t \to 1} \ln(t^t) = \lim_{t \to 1} t \ln(t) \text{.} \end{equation*}
Since \(\ln(t) \to 0\) as \(t \to 1\text{,}\) we may apply the multiplicativity rule of Pattern 23.2.2 to determine
\begin{equation*} \lim_{t \to 1} t \ln(t) = 1 \cdot 0 = 0 \text{.} \end{equation*}
Therefore, Pattern 23.6.18 tells us that
\begin{equation*} \lim_{t \to 1} t^t = e^0 = 1 \text{.} \end{equation*}
Now we turn back to indeterminate forms involving exponents.
Example 23.6.21.
Consider
\begin{equation*} \lim_{x \to 0^{+}} t^t \text{,} \end{equation*}
which is of \(0^0\) form. Applying Strategy 23.6.20, instead we compute
\begin{equation*} \lim_{t \to 0^{+}} \ln(t^t) = \lim_{t \to 0^{+}} t \ln(t) \text{.} \end{equation*}
This new limit is of \(0 \cdot (-\infty)\) form, so we apply Strategy 23.6.14:
\begin{align*} \lim_{t \to 0^{+}} t \ln(t) \amp = \lim_{t \to 0^{+}} \frac{\ln(t)}{1/t}\\ \amp \lhopeq \lim_{t \to 0^{+}} \frac{1/t}{-1/t^2} \\ \amp = \lim_{t \to 0^{+}} -t \\ \amp = 0 \text{.} \end{align*}
Completing strategy Strategy 23.6.20, we have
\begin{equation*} \lim_{t \to 0^{+}} t^t = e^0 = 1 \text{.} \end{equation*}
Example 23.6.22.
Consider
\begin{equation*} \lim_{t \to \infty} {(1 + t)}^{1/t} \text{,} \end{equation*}
which is of \(\infty^0\) form. Applying Strategy 23.6.20, instead we compute
\begin{equation*} \lim_{t \to \infty} \ln\bbrac{{(1 + t)}^{1/t}} = \lim_{t \to \infty} \frac{\ln(1 + t)}{t} \text{.} \end{equation*}
This new limit is of \(\infty / \infty\) form, so apply L’Hôpital’s Rule:
\begin{equation*} \lim_{t \to \infty} \frac{\ln(1 + t)}{t} \lhopeq \lim_{t \to \infty} \frac{1 / (1 + t)}{1} = 0 \text{.} \end{equation*}
Completing strategy Strategy 23.6.20, we have
\begin{equation*} \lim_{t \to \infty} {(1 + t)}^{1/t} = e^0 = 1 \text{.} \end{equation*}
Example 23.6.23. Alternative characterization of Euler’s number.
Consider
\begin{equation*} \lim_{t \to 0} {(1 + t)}^{1/t} \text{,} \end{equation*}
which is of \(1^\infty\) form. Applying Strategy 23.6.20, instead we compute
\begin{equation*} \lim_{t \to 0} \ln\bbrac{{(1 + t)}^{1/t}} = \lim_{t \to 0} \frac{\ln(1 + t)}{t} \text{.} \end{equation*}
This limit is now of \(0/0\) form, but this time we don’t need to apply L’Hôpital’s Rule. Instead, let’s trade variable \(t\) for \(\change{t}\text{:}\)
\begin{align*} \lim_{\change{t} \to 0} \frac{\ln(1 + \change{t})}{\change{t}} \amp = \lim_{\change{t} \to 0} \frac{\ln(1 + \change{t}) - 0}{\change{t}}\\ \amp = \lim_{\change{t} \to 0} \frac{\ln(1 + \change{t}) - \ln(1)}{\change{t}} \text{.} \end{align*}
This last limit is precisely the limit form of the derivative of the logarithm function at \(t = 1\) (see Fact 23.5.21). Therefore,
\begin{align*} \lim_{t \to 0} \ln\bbrac{{(1 + t)}^{1/t}} \amp = \lim_{\change{t} \to 0} \frac{\ln(1 + \change{t}) - \ln(1)}{\change{t}}\\ \amp = \funceval{\ddt{\bbrac{\ln(t)}}}{t = 1} \\ \amp = \funceval{\frac{1}{t}}{t = 1} \\ \amp = 1 \text{.} \end{align*}
From this we conclude
\begin{equation*} \lim_{t \to 0} {(1 + t)}^{1/t} = e^1 = e \text{.} \end{equation*}
Remark 23.6.24.
The limit in the preceding example is sometimes taken as the definition of Euler’s number, and then Definition 9.3.1 is proved as a consequence.

Subsection 23.6.3 Justification

As is the case with many concepts in calculus, averages are the key. Recall that Mean Value Theorem says that a continuous function always achieves its average value at least once on a specific domain:
\begin{equation*} f(c) = \avg{f} = \frac{1}{b - a} \ccmint{a}{b}{f(t)}{t} \end{equation*}
for at least one input \(c\) in the domain \(a \le t \le b\text{.}\) If we apply this fact to a derivative function, then we obtain the following version of the Mean Value Theorem.
Justification.
For simplicity, we will justify this under the extra assumptions that the derivative function \(f'(t)\) is defined and continuous on the closed interval \(a \le t \le b\text{.}\) In that case the integral version of the Mean Value Theorem applies, and we can say that there is at least one input value \(t = c\) in the domain where
\begin{equation*} f'(c) = \frac{1}{b - a} \ccmint{a}{b}{f'(t)}{t} \text{.} \end{equation*}
But \(f\) is an antiderivative of \(f'\text{,}\) so we simply apply Pattern 13.6.2 to obtain
\begin{equation*} \ccmint{a}{b}{f'(t)}{t} = f(b) - f(a) \text{.} \end{equation*}
Remark 23.6.26.
  • Note that the difference quotient
    \begin{equation*} \frac{f(b) - f(a)}{b - a} \end{equation*}
    is the average rate of variation of \(f\) over the domain \(a \le t \le b\) (see Definition 1.3.3), and the derivative value \(f'(c)\) represents the instantaneous rate of variation. So Theorem 23.6.25 really does represent the same information as the integral version of Mean Value Theorem: the differential form says that the rate-of-variation function \(f'\) acheives its average value at least once.
  • It is not to difficult to justify Theorem 23.6.25 without making the extra assumptions that \(f'(t)\) is defined and continuous on the closed interval \(a \le t \le b\text{.}\) First one considers the special case that \(f(b) = f(a)\text{,}\) and combines the Extreme Value Theorem with Pattern 18.2.10. Then, in the case \(f(b) \neq f(a)\text{,}\) one can reduce to the previous case by considering function \(d(t) = f(t) - \ell(t) \) where \(\ell(t)\) represents the secant line between the points \(\bbrac{a,f(a)}\) and \(\bbrac{b,f(b)}\text{.}\) We leave the details to you, the reader.
L’Hôpital’s Rule involves two functions in a ratio. If both the numerator function \(f\) and the denominator function \(g\) satisfy the hypotheses of the Differential form of the Mean Value Theorem, then there is some input \(t = c_1\) where \(f'(c_1)\) is equal to the average rate of \(f\) over the domain, and there is also some input \(t = c_2\) where \(g'(c_2)\) is equal to the average rate of \(g\) over the domain. Since \(f\) and \(g\) may be completely unrelated functions, there is no reason to think that these two input values are the same. But the following fact comes close to saying that we can take them to be the same input value.
Justification.
First we note that it is not possible for \(g(b) - g(a) = 0\) to be true, which is important because if it were then our conclusion would be impossible. The Differential form of the Mean Value Theorem tells us that there is some input value \(t = d\) where
\begin{equation*} g'(d) = \frac{g(b) - g(a)}{b - a} \text{.} \end{equation*}
If \(g(b) - g(a) = 0\) were true, then the above would say that \(g'(d) = 0\text{,}\) but we have assumed that \(g'(t)\) is never \(0\) on the domain in question.
We are looking for an input value \(t = c\) so that
\begin{equation*} \frac{f'(c)}{g'(c)} = \frac{F}{G} \end{equation*}
for
\begin{align*} F \amp = f(b) - f(a) \text{,} \amp G \amp = g(b) - g(a) \text{.} \end{align*}
(Note that \(F\) and \(G\) are specific numbers, not functions.)
Now consider \(h(t) = G \cdot f(t) - F \cdot g(t)\text{.}\) Since \(f\) and \(g\) satisfy the hypotheses of the Differential form of the Mean Value Theorem, so does \(h\text{,}\) and so there is a input value \(t = c\) where
\begin{equation*} h'(c) = \frac{h(b) - h(a)}{b - a} \text{.} \end{equation*}
But
\begin{align*} h(b) - h(a) \amp = \bbrac{G \cdot f(b) - F \cdot g(b)} - \bbrac{G \cdot f(a) - F \cdot g(a)} \\ \amp = G \cdot \bbrac{f(b) - f(a)} - F \cdot (g(b) - g(a)) \\ \amp = G \cdot F - F \cdot G \\ \amp = 0 \end{align*}
So, in fact,
\begin{gather*} h'(c) = \frac{0}{b - a} \\ h'(c) = 0 \\ G \cdot f'(c) - F \cdot g'(c) = 0 \\ G \cdot f'(c) = F \cdot g'(c) \\ G \cdot \frac{f'(c)}{g'(c)} = \frac{F}{G} \end{gather*}
as desired.
Now we are in a position to justify L’Hôpital’s Rule.
Outline of justification of L’Hôpital’s Rule for \(0/0\) form.
Assume \(f\) and \(g\) are functions that satisfy the three conditions in Pattern 23.6.4. As discussed in Remark 23.1.7, the value of
\begin{equation*} \lim_{t \to c} \frac{f(t)}{g(t)} \end{equation*}
does not depend on the values of \(f(c)\) and \(g(c)\text{,}\) or even if these values are defined. So without loss of generality we may assume
\begin{equation*} f(c) = g(c) = 0 \text{.} \end{equation*}
Using Fact 23.3.4, first let’s investigate the limit as \(t \to c^{+}\text{.}\) So consider an input sequence \(\nseq{t} \to c\) with \(t_n \gt c\) for all \(n\text{.}\) Since this sequence approaches \(c\text{,}\) we may assume that it is contained within the domains of continuity and differentiability of \(f\) and \(g\) in the assumptions in the statement of L’Hôpital’s Rule. This means that Cauchy’s Mean Value Theorem applies to every domain \(c \le t \le t_n\text{,}\) and so for each \(n\) there exists an input value \(t = k_n\) between \(c\) and \(t_n\) for which
\begin{equation*} \frac{f'(k_n)}{g'(k_n)} = \frac{f(t_n) - f(c)}{g(t_n) - g(c)} \text{.} \end{equation*}
But recall that we have assumed \(f(c) = g(c) = 0\text{,}\) so actually
\begin{equation*} \frac{f'(k_n)}{g'(k_n)} = \frac{f(t_n)}{g(t_n)} \text{.} \end{equation*}
So the sequences
\begin{align*} \amp \lrseq{\frac{f'(t_n)}{g'(t_n)}} \amp \amp \lrseq{\frac{f(k_n)}{g(k_n)}} \end{align*}
must both converge to the same limit or must both diverge in the same way. By the Sequence Squeeze Theorem, since \(c \le k_n \le t_n\) and \(\nseq{t} \to c\text{,}\) we also have \(\nseq{k} \to c\text{.}\) So if
\begin{equation*} \lim_{t \to c^{+}} \frac{f'(t)}{g'(t)} \end{equation*}
exists, it must have the same limit value as the sequence \(\bbseq{f'(k_n)/g'(n)}\text{,}\) which has the same limit value as the sequence \(\bbseq{f(t_n)/g(t_n)}\text{,}\) which has the same limit value as
\begin{equation*} \lim_{t \to c^{+}} \frac{f(t)}{g(t)} \text{.} \end{equation*}
Having demonstrated that
\begin{equation*} \lim_{t \to c^{+}} \frac{f(t)}{g(t)} = \lim_{t \to c^{+}} \frac{f'(t)}{g'(t)} \text{,} \end{equation*}
and recognizing that a similar argument could be used to demonstrate the same for \(t \to c^{-}\text{,}\) we have our desired conclusion.
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