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.
If you get these down, you’ll be able to easily solve almost any problem you’re given.For a discussion of why these approaches work, please read our blog post 0 Divided by 0: Solve Limit Problems in Calculus, Part 1.
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.
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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 not the 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. : )