## Evaluating Limits

### Evaluating Limits: Problems and Solutions

You probably already understand the basics of what limits are, and how to find one by looking at the graph of a function. So we’re going to jump right into where most students initially have some trouble: how to actually evaluate or *compute* a limit in homework and exam problems, especially in cases where you initially get 0 divided by 0. We’ll illustrate the six tactics you *must* know, and then let you practice each.

For a discussion of *why* these approaches work, please read our blog post 0 Divided by 0: Solve Limit Problems in Calculus, Part 1.

*x*into

*f(x)*. If you obtain a number (and in particular, if you don’t get $\dfrac{0}{0}$), you have your answer and are finished. In that case, these problems are completely straightforward.

**Example**.

Find $\displaystyle{\lim_{x \to 2}(x^3-5x + 7)}$.

*Solution*.

\begin{align*} \lim_{x \to 2}(x^3-5x + 7) &= (2)^3 -5(2) + 7 \\ &= 8 -10 + 7 = 5 \quad \cmark \end{align*}

Practice this (simple!) tactic in the next few problems. The solutions are immediately available using the **Click to View Calculus Solution** toggle. Substitution practice problem #3 illustrates an important point.

As the preceding problem illustrates, when you substitute the limit-value into the function, if you obtain \(\frac{\text{a non-zero number}}{0}\text{,} \), then you know immediately that the limit does not exist (DNE):

\[\text{If} \quad \lim_{x \to a}f(x) = \frac{\text{a non-zero number}}{0}\] \[\text{then the limit does not exist (DNE).}\]

**Example**.

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

*Solution.*

We first try substitution:

\begin{align*}

\lim_{x \to 2}\frac{x^2-4}{x-2} &= \frac{2^2 -4}{2 – 2} = \frac{0}{0}

\end{align*}

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:

$\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$ | \(=\) | $\displaystyle{\lim_{x \to 2}\frac{(x+2)(x-2)}{x-2}}$ | ||||

Ah, now we can cancel the problematic term: | ||||||

\(=\) | $\displaystyle{\lim_{x \to 2}\frac{(x+2)\cancel{(x-2)}}{\cancel{x-2}}}$ | |||||

\(=\) | $\displaystyle{\lim_{x \to 2} \,(x+2)}$ | |||||

And now easy Substitution to finish: | ||||||

\(=\) | $\displaystyle{2 + 2 = 4 \quad \cmark}$ |

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.

*Note*: Every Calculus exam on limits that we’ve ever seen has at *least* one problem that requires this tactic.

**Example**.

Find $\displaystyle{ \lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x}}$.

*Solution*.

As always, we first try Substitution:

$$\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} = \frac{\sqrt{5} – \sqrt{5}}{0} = \dfrac{0}{0}$$

Since this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We have more work to do.

So let’s try to rationalize the expression. The square-root part is $\sqrt{x+5} – \sqrt{5}$, so we multiply both the numerator and the denominator by the conjugate $\dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}}=1$:

\begin{align*}

\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} &= \lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} \cdot \dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}} \\[8px]
&= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \sqrt{x+5}\sqrt{5} – \sqrt{5}\sqrt{x+5} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px]
&= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \cancel{\sqrt{x+5}\sqrt{5}} – \cancel{\sqrt{5}\sqrt{x+5}} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px]
&= \lim_{x \to 0}\dfrac{(x+5) – 5}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px]
&= \lim_{x \to 0}\dfrac{x}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px]
&= \lim_{x \to 0}\dfrac{\cancel{x}}{\cancel{x}[\sqrt{x+5} + \sqrt{5}]} \\[8px]
&= \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} \\[8px]
&= \dfrac{1}{\sqrt{0+5} + \sqrt{5}} = \dfrac{1}{2\sqrt{5}} \quad \cmark

\end{align*}

Notice that when we multiplied by the conjugate, we multiplied out all of the terms in the numerator, because that’s how we get rid of the square root. But we *didn’t* multiply the terms in the denominator; instead we kept writing it as $x \left(\sqrt{x+5} + \sqrt{5} \right)$. That’s because a few steps later the *x* canceled.

Something similar will always happen, so in that early step don’t multiply out the part that you didn’t set out to rationalize. Instead just carry those terms along for a while, until you can cancel something.

1. Remember your fundamental trig identities. For instance:

- $\tan x = \dfrac{\sin x}{\cos x}$
- $\csc x = \dfrac{1}{\sin x}$
- $\sin^2(x) + \cos^2(x) = 1$

2. Early in the semester, there are two “Special Limits” you just have to memorize:

\begin{align*}

\text{I. } & \lim_{x \to 0}\frac{\sin(x)}{x} = 1 \\[16px]
\text{II. } & \lim_{x \to 0}\frac{1-\cos(x)}{x} = 0

\end{align*}

The following problems illustrate how you’ll frequently have to make use of these facts.

Want access to *all* of our Calculus problems and solutions? Buy full access now — it’s quick and easy!

\(\lim_{x \to 0}\left( 5x^5+4x^4-3x^3+2x^2-x+1\right)^3 \)

how can you solve this I don’t get it.

Hi, Adrian,

Thanks for asking! This is a problem that can use Tactic #1 above, Substitition. Remember, if you simply substitute in the value and get an answer that’s not 0/0, then you’re done. That works here:

\begin{align*}

\lim_{x \to 0}\left( 5x^5+4x^4-3x^3+2x^2-x+1\right)^3 &= \left( 5(0)^5 + 4(0)^4 -3(0)^3 + 2(0)^2 – 0 + 1\right)^3 \\[8px]

&= (0 + 0 – 0 + 0 -0 + 1)^3 \\[8px]

&= (1)^3 = 1 \quad \cmark

\end{align*}

We hope that helps. Please let us know if you have other questions!

If (1+x)^1/x is equal to e when

lim x approaches to 0, (1+x^2)^1/x^2, is it equal to e^2?

Hi, Nana,

Thanks for asking! We’re going to provide two answers here for reasons you’ll see. [You may need to reload this page for the math below to render correctly.]

First, the question doesn’t make sense to us since $\displaystyle{\lim_{x \to 0} \dfrac{1+x}{x} \ne e }.$ Instead, since

\[\frac{1+x}{x} = \frac{1}{x} + 1\]

and $\dfrac{1}{x}$ blows up as $x \rightarrow 0,$ the limit does not exist. The graph below also shows visually what happens near $x=0.$

\[\lim_{x \to 0} \left(\dfrac{1}{x} + 1 \right) \text{ does not exist} \quad \cmark \]

Since the “if” in your question isn’t correct, we can’t procede past that.

At the same time, we know that math can be difficult to type into a comment, and so you may not have written the equation you meant to. So let’s provide a more general answer to what you might be asking:

If $\displaystyle{\lim_{x \to 0} f(x) = L },$ does $\displaystyle{\lim_{x \to 0} [f(x)]^2 = L^2 }?$

Then the answer is yes. In general,

In general, $\displaystyle{\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a}f(x) \right]^n}$

where $n$ is a positive integer.

Note, however, that your second function as-written is

notthe square of the first function. If we take $f(x) = \dfrac{1+x}{x},$ then\[[f(x)]^2 = \dfrac{(1+x)^2}{x^2} = \dfrac{1 + 2x + x^2}{x^2} \ne \dfrac{1+x^2}{x^2}\]

and so we can’t draw any “if … then …” conclusions.

If you’d like to clarify the question, we’d be happy to try to provide a more useful answer. As it is, we hope some of this is helpful!

Thanks again for asking. : )

Lim (x^2+3x-10)/x^2+5

x=-5

Thanks for asking, Khalid! (You may need to refresh your browser to see the answer below correctly rendered.)

Simple substitution works in this case, since the denominator is non-zero:

\begin{align*}

\lim_{x \to -5}\frac{x^2 +3x -10}{x^2 + 5} &= \frac{(-5)^2 + 3(-5) – 10}{(-5)^2+5} \\[8px]

&= \frac{25 – 15 -10}{25 + 5} \\[8px]

&= \frac{0}{30} \\[8px]

&= 0 \quad \cmark

\end{align*}

Remember to always try substitution first. If it works, you’re done!

We hope that helps. : )