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.

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

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:

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

\[\lim_{x \to a} f(x) = f(a)\]

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 \[\lim_{x \to a} f(x) = L\]

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).

The following problems give you practice at finding limits using Substitution. We’ll then examine a few complications below.

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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) =

\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:

- 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

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

*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{-3}{3-3} &= \frac{-3}{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.

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

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

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

*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. . .

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 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.)

- 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! - 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. - If you find $f(a) = \dfrac{\text{non-zero number}}{0},$ then the limit does not exist (DNE).

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