Let’s look at substitution to find a limit, the most straightforward technique we have to find a function’s limit at a given point. You can of course practice using our problems with complete solutions.
This approach works when a function (1) is defined at the point of interest and (2) behaves smoothly near that point, meaning there are no jumps or gaps there. For example, let’s consider these six functions, each at a different point:
The box below contains an interactive Desmos calculator for each of the functions above, with that function’s particular point of interest marked. As you can see in each case, the limit simply equals the function’s value at that point: $\displaystyle{\lim_{x \to a}f(x) = f(a).}$
The key take-away:
As you saw in the examples above, when Substitution works you can find the limit easily, often in one line, which makes these problems quick. And we don’t want to over-complicate that.
is a number L (if one exists) such that $f(x)$ is as close to L as we want
whenever x is sufficiently close to a.
We saw earlier various cases where $\displaystyle{\lim_{x \to a} f(x) \ne f(a) }$ (figure (a), below). By contrast, for these “nice, smooth functions” the limit does simply equal $f(a)$ (figure (b) below).
The following problems give you practice at finding limits using Substitution. We’ll then examine a few complications below.
Let’s consider a small complication that arises when we use Substitution to find a one-sided limit. The following example illustrates.
Recall the piecewise function we used to introduce one-sided limits:
\[f(x) =
\begin{cases}
-3 & \text{for } x \lt 0 \\[8px]
\phantom{-}5 & \text{for } x \ge 0
\end{cases}\]
We imagine that at this stage, if we ask you for the value of $\displaystyle{\lim_{x \to 0^-} f(x)},$ you would answer . . .
Solution.
Clearly $\displaystyle{\lim_{x \to 0^-} f(x)} = -3 \quad \cmark$
But notice that $f(0) \ne -3;$ instead, $f(0) = 5.$ This question illustrates that we need to be careful when attempting to use Substitution at a point where a piecewise function changes from one “piece” to another, as happens here at $x = 0.$ Fortunately, you now deeply understand the difference between $\displaystyle{\lim_{x \to 0^-} f(x)}$ and $f(0),$ and so know that what we actually care about in finding the limit is the function’s value L that we can get as close to as we want when approaching $x=0$ from the left (since the limit specifies $x \to 0^-$). And that means we’re interested in the function’s expression for $x \lt 0,$ $f(x) = -3,$ even though this is not the expression that defines $f(0).$
As Example 1 illustrates, when finding a one-sided limit at a point where the function’s expression changes:
Because we’re looking at the limit as $x \to 2^+$, approaching $x=2$ from the right, we want the expression for $x \gt 2:$
\[ f(x) = -3x+5 \quad \text{for } x \gt 2\]
Then applying the method of Substitution yields
\begin{align*}
\lim_{x \to 2^{+} }f(x) &= -3(2) + 5 \\[8px]
&= -6 + 5 \\[8px]
&= -1 \implies \; \text{ (C) } \; \cmark
\end{align*}
We’re aproaching $x= -\dfrac{\pi}{2}$ from the right, and so focus on the expression for $-\dfrac{\pi}{2} \le x \le 1:$
\[\sin(4x) \quad \text{for } -\dfrac{\pi}{2} \le x \le 1 \quad \blacktriangleleft\]
Then
\begin{align*}
\lim_{x \to -\dfrac{\pi}{2}^+}f(x) &= \sin \left(4 \left(-\frac{\pi}{2}\right) \right) \\[8px]
&= \sin \left( -2\pi \right) \\[8px]
&= 0 \implies \quad\text{ (B)} \quad \cmark
\end{align*}Let’s consider another complication: sometimes when you use Substitution to compute $f(a),$ you obtain a zero in the denominator but not in the numerator, meaning $f(a)$ is undefined. The following example illustrates.
Find $\displaystyle{ \lim_{x \to 3} \dfrac{-5}{x-3}}.$
Solution.
Given our earlier work, you might immediately recognize that this function has a vertical asymptote at $x=3.$ But perhaps you didn’t recognize that (or let’s pretend that you didn’t), and so to find this limit you blithely try Substitution as we did in all of the problems above:
\begin{align*}
\lim_{x \to 3} \dfrac{-5}{x-3} \overset{?}{=} \frac{-5}{3-3} &= \frac{-5}{0} \\[8px]
&= \text{undefined}
\end{align*}
The fact that you obtain “undefined” there does not automatically mean that the limit doesn’t exist. Instead, it means we need to think more deeply: Substitution has failed as a tactic, and so we have more work to do.
Specifically, once you obtain that particular result, you might now think: “Oh, right. The numerator is non-zero while the denominator goes to zero as $x \to 3,$ and so the function blows up at $x=3.$ There must be a vertical asymptote there, and the limit does not exist (DNE).” And that reasoning is correct:
\[\lim_{x \to 3} \dfrac{-5}{x-3} = \text{ DNE} \quad \cmark\]
We’re providing the graph to illustrate the conclusion, but again encourage you to make such thinking part of your repertoire so you don’t have to graph every function to be able to reason correctly. Indeed, you are likely to encounter exam questions where you are not allowed to use a graphing calculator, and so such reasoning is required.
As the preceding example illustrates, when you substitute the limit-value into the function, if you obtain
\(\frac{\text{a non-zero number}}{0} \), then you know immediately that the limit does not exist (DNE):
\[\text{If} \quad \lim_{x \to a}f(x) \overset{?}{=} f(a) = \frac{\text{a non-zero number}}{0}\text{,}\]
\[\text{then the limit does not exist (DNE).}\]
As the next problem illustrates, while we may quickly conclude that the limit does not exist, the answer choices may require a bit of further reasoning.
Since both the numerator and the denominator are squared-terms, we are guaranteed that the function always returns a positive value (except at $x=0,$ where $f(0) = 0$). Hence as $x \to -1$ and the denominator tends toward zero, the function blows up in the positive direction: it grows and Grows and GROWS becoming more and MORE positive. And we can make the function’s output as large a positive number as we would like by being sufficiently close to $x=-1.$ Hence
\[\lim_{x \to -1} \dfrac{x^2}{(x+1)^2} = \infty \quad \cmark\]
We hope the graphical image that popped into your head was roughly the one above. If not, keep going: the only way to develop this “create rough-graph in my head” skill is to keep doing problems.
As usual, we first try Substitution. Recall that $\ln(1) = 0.$ And while we don’t know offhand what the value of $\cos(1)$ is, we do know that it’s not zero. So we’ll just call it “$\cos(1)$.”
\begin{align*}
\lim_{x \to 1} \frac{\cos(x)}{\ln(x)} &\overset{?}{=} \frac{\cos(1)}{0} \\[8px]
&= \dfrac{\text{non-zero number}}{0} \implies \quad\text{ (E) DNE} \quad \cmark
\end{align*}Let’s consider now an example where substitution gives the result $f(a) = \dfrac{0}{0}.$
Find $\displaystyle{\lim_{x \to 2} \dfrac{x^2-4}{x-2}}.$
Solution.
We begin, as usual, by trying Substitution:
\[ \lim_{x \to 2}\frac{x^2-4}{x-2} = \frac{4-4}{2-2} = \frac{0}{0} \]
Hmmm. We’ve seen that if we have $\dfrac{\text{non-zero number}}{0},$ then we know the limit does not exist. But what if we have $\dfrac{0}{0}$??
That’s a different challenge. Let’s pause this example for a moment. . .
Specifically, the $\dfrac{0}{0}$ result signals that need to use a different method to find the limit. Fortunately, three simple tactics will let you solve most problems: (1) factor; (2) rationalize; (3) use algebra.
We’ll look at the first, factoring, on the next screen. Before we do, we’ve seen the limit in the preceding example before. In fact we used that function to introduce limits. Do you remember what the graph of that function looked like, and how we quickly arrived at that graph? (No worries if not; we’re about to revisit the key reasoning and then extend it.)
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