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B.2 Substitution to find a limit

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:

  • a linear function: $-4x + 2$ at $x=6$;
  • a polynomial: $x^3 – 5x + 7$ at $x=-1$;
  • a rational function (a quotient of two polynomials): $\dfrac{x^3-1}{x^5 – 2x^3 +1}$ at $x=2$;
  • an exponential function: $e^{-x}$ at $x=0.5$;
  • a logarithmic function: $\ln(x^2 + 1)$ at $x=0$;
  • a trig function: $\sin(2x)$ at $x=\pi/4.$

In each of these cases you can imagine a rough picture of the function’s graph, and then see in your mind’s eye how the value of the limit at the point of interest is simply equal to the function’s value at that 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:

Substitution to find a limit
When finding the limit $\displaystyle{\lim_{x \to a}f(x) },$ if you substitute $x=a$ into $f(x)$ and obtain a number for $f(a)$
(and don’t get “undefined”), then the limit is simply that value $f(a)$:
\[\lim_{x \to a} f(x) = f(a)\]

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.

Remember: a function’s value at a point and its limit at that point are distinct quantities
At the same time, we want to emphasize again that $\displaystyle{\lim_{x \to a}f(x) }$ and $f(a)$ are two distinct quantities. Recall our working definition of limits:
\[\lim_{x \to a} f(x) = L\]

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).
When using substitution to find a limit, keep in mind the important distinction illustrated by these two figures: (a) Graph with a hole in it, such that f(2) = 5, which does not fall on the line that defines the function everywhere else. (b) A nice smooth parabola, such that every point of the function falls on the curve.
The following problems give you practice at finding limits using Substitution. We’ll then examine a few complications below.

Initial practice of substitution to find a limit

Practice Problem #1
Find $\displaystyle{ \lim_{x \to -2}\left(x^3 +4\right) }$. \begin{array}{lllll} \text{(A) }-4 && \text{(B) }12 && \text{(C) }-12 && \text{(D) }-8 && \text{(E) None of these} \end{array}
Show/Hide Solution
\begin{align*} \lim_{x \to -2}\left(x^3 +4\right) &= (-2)^3 +4 \\[4px] &= -8 +4 \\[4px] &= -4 \implies \quad\text{ (A)} \quad \cmark \end{align*}
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Practice Problem #2
Find $\displaystyle{\lim_{x \to 4}\sqrt{9+\sqrt{4x}}}$. \begin{array}{lllll} \text{(A) }\sqrt{13} && \text{(B) }\sqrt{11} && \text{(C) }5 && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
\begin{align*} \lim_{x \to 4}\sqrt{9+\sqrt{4x}} &=\sqrt{9+\sqrt{4(4) }} \\[4px] &= \sqrt{9+\sqrt{16}} = \sqrt{9+4} \\[4px] &= \sqrt{13} \implies \quad\text{ (A)} \quad \cmark \end{align*}
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Practice Problem #3
Find $\displaystyle{ \lim_{x \to 1} }\dfrac{\ln x}{x}$. \begin{array}{lllll} \text{(A) }1 && \text{(B) }e && \text{(C) }0 && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
\begin{align*} \lim_{x \to 1}\dfrac{\ln x}{x} &= \dfrac{\ln (1) }{1} \\[12px] &= \frac{0}{1} \\[12px] &= 0 \implies \quad\text{ (C)} \quad \cmark \end{align*}
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Practice Problem #4
Find $\displaystyle{ \lim_{x \to \pi/2} } \cos x$. \begin{array}{lllll} \text{(A) }1 && \text{(B) }0 && \text{(C) }\dfrac{1}{\sqrt{2}} && \text{(D) DNE } && \text{(E) None of these} \end{array}
Show/Hide Solution
Recall that $\cos \left(\dfrac{\pi}{2} \right) = 0$. Hence \begin{align*} \lim_{x \to \pi/2}\cos x &= \cos \left(\dfrac{\pi}{2} \right) \\[8px] &= 0 \implies \quad\text{ (B)} \quad \cmark \end{align*}
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Substitution and one-sided limits

Let’s consider a small complication that arises when we use Substitution to find a one-sided limit. The following example illustrates.

Example 1: One-Sided Limit

Recall the piecewise function we used to introduce one-sided limits:
\[f(x) =
-3 & \text{for } x \lt 0 \\[8px] \phantom{-}5 & \text{for } x \ge 0
\end{cases}\] We imagine that at this stage, if we ask you for the value of $\displaystyle{\lim_{x \to 0^-} f(x)},$ you would answer . . .
Graph of piecewise function f(x), which equals -5 for x less than zero, and +5 for x greater than or equal to 0.

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:

  • if you’re approaching $x=a$ from the left ($x \to a^-$), then use the expression appropriate for $x \lt a.$
  • if you’re approaching $x=a$ from the right ($x \to a^+$), then use the expression appropriate for $x \gt a.$

We don’t intend those bullet points as facts to memorize. Instead, they’re just a reminder to keep the meaning of limits in mind when you answer these types of questions, rather than just mindlessly substituting.
Practice Problem #5
Consider the following piecewise function: \begin{align*} f(x) &= \begin{cases} x^2 & \; x \le 2 \\[8px] -3x+5 & \; x \gt 2 \\[8px] \end{cases} \\[8px] \end{align*} Find $\displaystyle{ \lim_{x \to 2^{+}} f(x)}$. \begin{array}{lllll} \text{(A) }4 && \text{(B) }5 && \text{(C) }-1 && \text{(D) Does not exist} && \text{(E) None of these} \end{array}
Show/Hide Solution
Graph shows a parabola based at the origin ending on a closed-circle at (2, 4). The graph then jumps down to an open circle at (2, -1) and a downward-sloped line from there. A red arrow points toward x=2 from the right, toward the point (2, -1), which has an open-circle on it. 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*}
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Practice Problem #6
Consider the following piecewise function: \[f(x) = \begin{cases} -2x +6 & \text{for } x \lt -\dfrac{\pi}{2} \\[8px] \sin(4x) & \text{for } -\dfrac{\pi}{2} \le x \le 1 \\[8px] x^3 & \text{for } x \gt 1 \end{cases}\] Find $\displaystyle{\lim_{x \to -\dfrac{\pi}{2}^+}f(x) }.$ \begin{array}{lllll} \text{(A) }\pi + 6 && \text{(B) }0 && \text{(C) }1 && \text{(D) }-\dfrac{\pi^3}{8} && \text{(E) None of these} \end{array}
Show/Hide Solution
Graph of the piecewise function, with a red arrow pointing from the right toward the left at x = -pi/2, making it clear we want the portion of the function defined by the sin function. 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*}
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What about when $f(a) = \dfrac{\text{non-zero number}}{0},$ and so is undefined?

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.

Example 2: $f(a) = \dfrac{\text{non-zero number}}{0}$

Find $\displaystyle{ \lim_{x \to 3} \dfrac{-5}{x-3}}.$

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:
\lim_{x \to 3} \dfrac{-5}{x-3} \overset{?}{=} \frac{-5}{3-3} &= \frac{-5}{0} \\[8px] &= \text{undefined}
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.

Hyperbola centered at x=3, and a vertical red line showing the asymptote x=3.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.

Practice Problem #7
Answer without using a graphing calculator: $\displaystyle{\lim_{x \to -1} \dfrac{x^2}{(x+1)^2} = }$ \begin{array}{lllll} \text{(A) }-\dfrac{1}{4} && \text{(B) }1 && \text{(C) }\dfrac{1}{4} && \text{(D) }\infty && \text{(E) }-\infty \end{array}
Show/Hide Solution
If we try Substitution, we obtain \begin{align*} \lim_{x \to -1} \dfrac{x^2}{(x+1)^2} \overset{?}{=} \frac{(-1)^2}{(-1 + 1)^2} &= \frac{1}{0} \\[8px] &= \dfrac{\text{non-zero number}}{0} \implies \quad\text{DNE} \end{align*} If one of the answer choices was “DNE,” that would be correct and we’d be done. But “DNE” is not an answer choice. We do know that the choice must be (D) or (E), simply providing more information about the way in which the limit does not exist. Let’s think a bit further to determine which of those choices is correct:

Graph of the function showing that it goes to positive infinity as x approaches -1.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.
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Practice Problem #8
$\displaystyle{\lim_{x \to 1} \dfrac{\cos(x)}{\ln(x)}}$ = \begin{array}{lllll} \text{(A) }0 && \text{(B) }0.99985 && \text{(C) }1 && \text{(D) }-1 && \text{(E) DNE} \end{array}
Show/Hide Solution
Graph of the function showing the vertical asymptote at x = -1.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*}
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Finally, what about when $f(a) = \dfrac{0}{0},$ and so is undefined?

Let’s consider now an example where substitution gives the result $f(a) = \dfrac{0}{0}.$

Example 3: $f(a) = \dfrac{0}{0}$ (?!)

Find $\displaystyle{\lim_{x \to 2} \dfrac{x^2-4}{x-2}}.$

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. . .

In nearly all of your homework and test questions, when you try Substitution you’ll obtain 0 divided by 0. You then need another tactic to find the limit.
The wrinkle: We wouldn’t need the concept of the limit if you could always just substitute $x=a$ and find the function’s value at the point of interest. Instead, the truth is that when you try Substitution with nearly all of your homework and test questions, you’ll obtain $\dfrac{0}{0}$, “zero divided by zero.” That result is known as an indeterminate limit, which is a fancy way of saying “not yet known.” This “result” tells you that the actual answer could be anything — you just don’t know yet — and so you have more work to do.

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.)

The Upshot.

  1. When finding the limit $\displaystyle{\lim_{x \to a}f(x) },$ if you substitute $x=a$ into $f(x)$ and obtain a
    number for $f(a)$ (and don’t get “undefined”), then the limit is simply that value $f(a)$:
    \[\lim_{x \to a} f(x) = f(a)\] Hence, your first step in computing any limit should be to try Substitution. If it works and you obtain a number, you’re done!
  2. When finding a one-sided limit at an x-value where the function’s expression changes, be careful to choose the correct expression depending on whether you are finding the left-hand or right-hand limit.
  3. If you find $f(a) = \dfrac{\text{non-zero number}}{0},$ then the limit does not exist (DNE).

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