B.3 Factor to Find a Limit

Calculus Home Limits & Continuity B. Calculating Limits B.3 Factor to Find a Limit

Let’s see how we can frequently factor to find a limit. This is the tactic you will use most often to compute limits, so practice using our problems with complete solutions.

We’ll begin by continuing with the example we introduced at the bottom of the previous screen, and see how we can use the simple algebraic tactic of factoring to find the limit we’re after.

Example 1: Factor to find $\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$

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

Solution.
Step 1. Try Substitution.
We always try Substitution first, as we did at the bottom of the preceding screen:
\begin{align*}
\lim_{x \to 2}\frac{x^2-4}{x-2} &\overset{?}{=} \frac{4-4}{2-2} \\[8px] &= \frac{0}{0} \implies \text{indeterminate result}
\end{align*}
where “indeterminate result” means “we don’t know what the value actually is, and hence have more work to do.”

Step 2. Factor and simplify:
Let’s factor the numerator of our function:
\begin{align*}
\lim_{x \to 2}\frac{x^2-4}{x-2} &= \lim_{x \to 2}\frac{(x+2)(x-2)}{x-2}
\end{align*}
Ah, now we can cancel the problematic term:
\begin{align*}
\phantom{ \lim_{x \to 2}\frac{x^2-4}{x-2}} &= \lim_{x \to 2}\frac{(x+2)\cancel{(x-2)}}{\cancel{x-2}} \\[8px] &= \lim_{x \to 2} \,(x+2) \\[8px] \end{align*}
Step 3. Now easy Substitution to finish:
\begin{align*}
\phantom{ \lim_{x \to 2}\frac{x^2-4}{x-2}} &= 2 + 2 = 4 \quad \cmark
\end{align*}

Why does this approach work?
Notice that the function $\dfrac{x^2-4}{x-2}$ simplified to $x+2$ when we factored it. The only difference between the two functions is that $\dfrac{x^2-4}{x-2}$ isn’t defined for $x=2$, since the denominator is zero there, whereas $x+2$ is defined everywhere. To illustrate that $\dfrac{x^2-4}{x-2}$ is undefined at $x=2$, we show a hole in its graph at that point in the left-hand figure; at every other point, the graph is exactly the same as the the graph of $f(x) = x+2$, shown in the right-hand figure.

The left-hand graph, of (x^2-4)/(x-2), has a hole in it at x=2, since the function is undefined there. The right-hand graph of x+2 does not, since it is defined everywhere.
By comparing the two graphs, you can see why the limits are the same: whether or not the function is defined at $x=2,$ as we approach that value we get closer and closer to the height $y=4.$ The whole concept of the limit was created for just this situation, so we can imagine getting closer and closer to $x=2$ from the left or from the right without ever fully reaching that point. The process of factoring simply lets us rewrite the function in a form to which we can apply Substitution and conclude that
$$\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}} = \lim_{x \to 2}(x+2) = 4$$
[If you’d like to revisit the interactive Desmos graph for this limit, it’s the very first one we looked at in our Limits Introduction.]

As you will see in many, many problems to come, we will use this basic technique often: through algebraic
manipulation, we are able to rewrite a function that isn’t defined at the point of interest and “turn it into” a different form that is defined at the point of interest. Simple substitution then works to know the value of the function at that point, which in turn is the limit of the function at that point.

The rewritten function is not the same as the initial function

There is an important subtlety here: the two functions in the graphs above are not the same because they have different domains: the original function, $\dfrac{x^2-4}{x-2},$ is not defined at $x=2,$ whereas our rewritten function is. At every point other than $x=2,$ they are identical, but because they behave differently at this one point they are not identical functions. We can make them identical, however, by simply excluding this one point:
\[\frac{x^2-4}{x-2} = x+2 \quad \text{for } x \ne 2 \] And because we explicitly don’t care about what’s happening at $x=2,$ and instead only what happens close to $x=2,$ we have
\[\lim_{x \to 2}\frac{x^2-4}{x-2} =\lim_{x \to 2}(x+2) = 4\] More generally, we have the incredibly helpful rule:

Rewriting functions to find a limit

If $f(x) = g(x)$ when $x \ne a,$ then $\displaystyle{\lim_{x \to a}f(x) = \lim_{x \to a}g(x)},$ assuming the limits exist.

The rule merely encapsulates the reasoning we used above, and will use time and again.

This is almost the only time we’ll discuss this subtlety. But please keep in mind that the algebraic manipulations you’ll do in almost every problem allow us to find the limit of the original function only because of the way the limit is defined to focus on the behavior of the function close to, rather than at, the point of interest.

You saw in Example 1 above the three simple steps for the tactic of factoring to find a limit:

PROBLEM-SOLVING TACTIC: Factor to find a limit

Try Substitution. As we saw on the preceding screen, if it works as a tactic, you’re done. If, by contrast, you obtain $\dfrac{0}{0},$ then …
Factor and simplify. If you can factor the numerator and/or denominator, do so. You’ll almost always find that the problematic term in the denominator cancels, and you’ll be left with a new function for which you can use. . .
Substitution to finish.

Practice the tactic of factoring to find the limit in the next few problems. These are straightforward once you learn to recognize what to do.

Practice Problems: Factor to Find a Limit

Every Calculus exam that we’ve ever seen that includes limits as a topic has at least one problem that requires this tactic. Practice accordingly to be prepared.

Factor to Find a Limit: Scaffolded Exercise #1

Find $\displaystyle{ \lim_{x \to 1} }\,\frac{x-1}{x^2-1}.$

Solution.
Step 1. First try substitution:

Show/Hide Step 1 Solution

\[ \lim_{x \to 1} \,\frac{x-1}{x^2-1} = \dfrac{1-1}{1-1}=\dfrac{0}{0} \] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We thus have more work to do. So on to Step 2…

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Step 2. Factor and simplify:

Show/Hide Step 2 Solution

\begin{align*}
\lim_{x \to 1} \,\frac{x-1}{x^2-1} &= \lim_{x \to 1} \,\frac{x-1}{(x-1)(x+1)} \\[8px] &= \lim_{x \to 1} \,\frac{\cancel{x-1}}{\cancel{(x-1)}(x+1)} \\[8px] &= \lim_{x \to 1} \, \frac{1}{x+1}
\end{align*}

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Step 3. Substitution to finish:

Show/Hide Step 3 Solution

\begin{align*}
\phantom{ \lim_{x \to 1} \,\frac{x-1}{x^2-1}} &= \lim_{x \to 1} \, \frac{1}{x+1} \\[8px] &= \frac{1}{1+1} \\[8px] &= \frac{1}{2} \quad \cmark
\end{align*}

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Factor to Find a Limit: Scaffolded Exercise #2

Find $\displaystyle{ \lim_{x \to 3} }\,\frac{x^2-2x -3}{x-3}.$

Solution.
Step 1. Try substitution:

Show/Hide Step 1 Solution

\[ \lim_{x \to 3} \,\frac{x^2-2x -3}{x-3} = \dfrac{9-6-3}{3-3}=\dfrac{0}{0} \] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We thus have more work to do.

[collapse]

Step 2. Factor and simplify:

Show/Hide Step 2 Solution

\begin{align*}
\lim_{x \to 3} \,\frac{x^2-2x -3}{x-3} &= \lim_{x \to 3} \,\frac{(x+1)(x-3)}{x-3} \\[8px] &= \lim_{x \to 3} \,\frac{(x+1)\cancel{(x-3)}}{\cancel{(x-3)}} \\[8px] &= \lim_{x \to 3} \, (x+1)
\end{align*}

[collapse]

Step 3. Substitution to finish:

Show/Hide Step 3 Solution

\begin{align*}
\phantom{ \lim_{x \to 3}\frac{x^2-3}{x-3}} &= \lim_{x \to 3} \, (x+1) \\[8px] &= 3+1 \\[8px] &= 4 \quad \cmark
\end{align*}

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Practice Problem #1

Find $\displaystyle{ \lim_{x \to 5} } \, \frac{x-5}{x^2-25}$. \begin{array}{lllll} \text{(A) }10 && \text{(B) }\dfrac{1}{10} && -\text{(C) }-\dfrac{1}{10} && \text{(D) 0} && \text{(E) None of these} \end{array}

Show/Hide Solution

We first try substitution: \[\lim_{x \to 5} \frac{x-5}{x^2-25} \overset{?}{=} \frac{5-5}{5^2 -25} = \frac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So let’s factor the denominator: \begin{align*} \lim_{x \to 5} \frac{x-5}{x^2-25} &= \lim_{x \to 5} \frac{x-5}{(x+5)(x-5)} \\[4px] &= \lim_{x \to 5} \frac{\cancel{x-5}}{(x+5)\cancel{(x-5)}} \\[4px] &= \lim_{x \to 5}\frac{1}{x+5} \\[4px] &= \frac{1}{5+5} = \frac{1}{10} \implies \quad\text{ (B)} \quad \cmark \end{align*}

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Practice Problem #2

Find $\displaystyle{ \lim_{x \to -3} }\frac{x^2 -9}{x+3}$. \begin{array}{lllll} \text{(A) }6 && \text{(B) }-3 && \text{(C) }-6 && \text{(D) } 3 && \text{(E) nonexistent} \end{array}

Show/Hide Solution

We first try substitution: \[\lim_{x \to -3}\frac{x^2 -9}{x+3} \overset{?}{=} \dfrac{(-3)^2 – 9}{(-3) +3} = \dfrac{0}{0} \] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So let’s factor the numerator: \begin{align*} \lim_{x \to -3}\frac{x^2 -9}{x+3} &= \lim_{x \to -3}\frac{(x+3) (x-3)}{(x+3)} \\[8px] &= \lim_{x \to -3} \frac{\cancel{(x+3)}(x-3)}{\cancel{(x+3)}} \\[8px] &= \lim_{x \to -3} (x-3) \\[8px] &= (-3) -3 = -6 \quad \text{(C)} \quad \cmark \end{align*}

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Practice Problem #3

Find $\displaystyle{ \lim_{x \to 1} }\frac{x^2 + 4x -5}{x^2 - 1}$. \begin{array}{lllll} \text{(A) }6 && \text{(B) }3 && \text{(C) }-3 && \text{(D) }-6 && \text{(E) None of these} \end{array}

Show/Hide Solution

We first try substitution: \[\lim_{x \to 1} \frac{x^2 + 4x -5}{x^2 – 1} \overset{?}{=} \frac{1 +4(1) – 5}{1-1} = \frac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So let’s factor the numerator and the denominator: \begin{align*} \lim_{x \to 1} \frac{x^2 + 4x -5}{x^2 – 1} &= \lim_{x \to 1} \frac{(x+5)(x-1)}{(x-1)(x+1)}\\[8px] &= \lim_{x \to 1} \frac{(x+5)\cancel{(x-1)}}{\cancel{(x-1)}(x+1)} \\[8px] &= \lim_{x \to 1}\frac{x+5}{x+1} \\[8px] &= \frac{1+5}{1+1} = \frac{6}{2} \\[8px] &=3 \implies \quad\text{ (B)} \quad \cmark \end{align*}

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Practice Problem #4

Find $\displaystyle{ \lim_{x \to 0} }\,\frac{3x^3 -4x}{7x^3 +5x}$. \begin{array}{lllll} \text{(A) }-\dfrac{4}{5} && \text{(B) }-\dfrac{1}{12} && \text{(C) }\dfrac{3}{7} && \text{(D) DNE} && \text{(E) None of these} \end{array}

Show/Hide Solution

We first try substitution: \[\lim_{x \to 0} \frac{3x^3 -4x}{7x^3 +5x} \overset{?}{=} \frac{0 – 0}{0+0} = \frac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We notice that every term has an x in it, so let’s factor that out: \begin{align*} \lim_{x \to 0} \,\frac{3x^3 -4x}{7x^3 +5x} &= \lim_{x \to 0}\, \frac{x(3x^2 -4)}{x(7x^2+5)} \\[8px] &= \lim_{x \to 0} \frac{\cancel{x}(3x^2 -4)}{\cancel{x}(7x^2+5)} \\[8px] &= \lim_{x \to 0} \frac{(3x^2 -4)}{(7x^2+5)} \\[8px] &= \frac{(0 -4)}{(0+5)} \\[8px] &= -\frac{4}{5} \implies \quad\text{ (A)} \quad \cmark \end{align*}

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Practice Problem #5

Find $\displaystyle{ \lim_{x \to 1/3} }\,\dfrac{3x - 1}{1 - 9x^2}$. \begin{array}{lllll} \text{(A) }-\dfrac{1}{3} && \text{(B) }\dfrac{1}{2} && \text{(C) }-\dfrac{1}{2} && \text{(D) DNE} && \text{(E) None of these} \end{array}

Show/Hide Solution

We first try substitution: \[\lim_{x \to 1/3} \,\dfrac{3x – 1}{1 – 9x^2} \overset{?}{=} \dfrac{3\left(\dfrac{1}{3} \right) -1 }{1 – 9 \left(\dfrac{1}{3} \right)^2} = \dfrac{1-1}{1-1} = \dfrac{0}{0}\] Because this limit is in the form $\dfrac{0}{0}$, it is indeterminiate—we don’t yet know what it is. So let’s factor the denominator: \begin{align*} \lim_{x \to 1/3}\, \dfrac{3x – 1}{1 – 9x^2} &= \lim_{x \to 1/3} \dfrac{3x – 1}{(1 -3x)(1+3x)} \\[8px] &= \lim_{x \to 1/3}\, \dfrac{-(1-3x)}{(1 -3x)(1+3x)} \\[8px] &= \lim_{x \to 1/3} \dfrac{-\cancel{(1-3x)}}{\cancel{(1 -3x)}(1+3x)} \\[8px] &= \lim_{x \to 1/3}\dfrac{-1}{1+3x} \\[8px] &= \frac{-1}{1+3\left(\dfrac{1}{3} \right) } = \frac{-1}{1+1} \\[8px] &=-\dfrac{1}{2} \implies \quad\text{ (C)} \quad \cmark \end{align*}

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Practice Problem #6

Find $\displaystyle{\lim_{x \to 2} }\dfrac{x-2}{x^3 - 8}$.

Hint: Factor a cubic

Notice $x^3 – 8 = x^3 – 2^3.$ Then recall \[a^3 – b^3 = (a-b)\left(a^2 + ab + b^2 \right)\] You can multiply out the right hand side of the equation to verify the equality.

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\begin{array}{lllll} \text{(A) }\dfrac{1}{16} && \text{(B) }\dfrac{1}{8} && \text{(C) }\dfrac{1}{4} && \text{(D) }\dfrac{1}{12} && \text{(E) None of these} \end{array}

Show/Hide Solution

We first try substitution: \[\lim_{x \to 2} \dfrac{x-2}{x^3 – 8} \overset{?}{=} \dfrac{2-2}{2^3-8} = \dfrac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So we factor the denominator: \begin{align*} \lim_{x \to 2} \dfrac{x-2}{x^3 – 8} &= \lim_{x \to 2} \dfrac{x-2}{(x-2)(x^2 + 2x + 4)} \\[8px] &= \lim_{x \to 2} \dfrac{\cancel{x-2}}{\cancel{(x-2)}(x^2 + 2x + 4)} \\[8px] &= \lim_{x \to 2} \dfrac{1}{(x^2 + 2x + 4)}\\[8px] &= \frac{1}{2^2 + 2(2)+4} = \frac{1}{4 + 4 + 4} \\[8px] &= \frac{1}{12} \implies \quad\text{ (D)} \quad \cmark \end{align*}

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The Upshot

If when you try Substitution you obtain $\dfrac{0}{0},$ then if possible factor the numerator and/or denominator, and reduce the fraction. Almost always, Substitution will then work.

On the next screen we’ll explain another tactic you’ll need often to compute a limit, using conjugates.

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Anonymous

8 months ago

For Question 6: the answer is (1/4) because when you multiply out the
expression (x-2)^2 you get (x^2-4x+4) and then plug in 2 in places of
x which gives you 1/4.

Last edited 8 months ago by Matheno

Matheno

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Reply to Anonymous

8 months ago

Thanks for writing to let us know you think you spotted an error! As it turns out our solution is correct, as you can see from the graph. Our algebra is also correct.

I’m guessing that you made a common, and quite understandable, error thinking that
\[x^3 – 8 \overbrace{=}^? (x-2)(x-2)^2\] but if you multiply out the right-hand side you’ll see that it does not equal the left-hand side. Instead, the formula we have in the solution is correct.

Or did you factor another way that led you to your conclusion? If so, please let us know and we’ll be happy to think that through with you.

For now, thanks again for writing in! And please let us know if you spot anything else you think is an error or have any other questions. We’re happy to help!

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