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3 Common Limit Problems You Must Know How to Solve

Are you having trouble solving Calculus limit problems, even though you understand the concept? In this post we explain three approaches you’ll use again and again, especially in problems where you initially get “0/0.”

Update: As of September 2022, we have much more interactive ways for you to learn about the foundational concept of Limits, making heavy use of Desmos graphing calculators. Please visit our Limits Chapter to really get this material down for yourself. It’s all free, and waiting for you!

I. Factoring

You’ll use factoring most often.
You’ll use this approach most often. For example, consider the problem
$$\lim_{x \to 4}\dfrac{x^2 – 16}{x-4} = ?$$
If you just plug in $x = 4$, you find
\lim_{x \to 4}\dfrac{x^2 – 16}{x-4} = \dfrac{16 – 16}{4-4} = \dfrac{“0”}{0}
Open to read why we put quotes around that result.
We put quotes around $\dfrac{“0”}{0}$ because we can’t actually have 0 in the denominator; that’s undefined. Indeed, the function $\dfrac{x^2 – 16}{x-4}$ is undefined at $x = 4$ for exactly this reason. We still need to find this limit, though, which tells us what value the function approaches as we get closer and closer to $x = 4$, without ever actually reaching 4.
When you get $\dfrac{“0”}{0}$, the first thing you should try is factoring the numerator or the denominator.
Continuing with our example, let’s factor the numerator:

\lim_{x \to 4}\dfrac{x^2 – 16}{x-4} &= \lim_{x \to 4}\dfrac{(x+4)(x-4)}{x-4} \\ \\
&= \lim_{x \to 4}\dfrac{(x+4)\cancel{(x-4)}}{\cancel{x-4}} \\ \\
&= \lim_{x \to 4}[x + 4] \\ \\
&= 8 \quad \cmark

We guarantee that if you can factor the numerator or the denominator, the problematic “0” term in the denominator will cancel like it did here. At that point, you can just plug in for x, and be done.

Open to see two more example problems that use factoring.
1. In this one you factor the denominator.
\lim_{x \to -3}\dfrac{x+3}{x^2 – 9} &= \text{ ?} \\ \\
&= \lim_{x \to -3}\dfrac{x+3}{(x+3)(x-3)} \\ \\
&= \lim_{x \to -3} \dfrac{\cancel{x+3}}{\cancel{(x+3)}(x-3)} \\ \\
&= \lim_{x \to -3}\dfrac{1}{x – 3}\\ \\
&= -\dfrac{1}{6} \quad \cmark

2. Factoring this one is just a little harder, but the idea is exactly the same.
\lim_{x \to 5}\dfrac{x^2 – 4x – 5}{x – 5} &= \text{ ?} \\ \\
&= \lim_{x \to 5}\dfrac{(x-5)(x+1)}{x – 5} \\ \\
&= \lim_{x \to 5}\dfrac{\cancel{(x-5)}(x+1)}{\cancel{x – 5}} \\ \\
&= \lim_{x \to 5}[x + 1] \\ \\
&= 6 \quad \cmark\end{align*}

The upshot: Factor whenever you can.


  • Our Chapter on Limits (everything you need to know, with interactive components to help you develop a feel for limits, and many problem solving tactics you can practice)

II. Expand the Polynomial

You probably have some problems that look like
$$\lim_{h \to 0} \dfrac{\text{stuff in the numerator}}{h} = ?$$
Again if you just plug in $h = 0$, you get that problematic “0” in the denominator.

Usually in these problems you aren’t able factor the numerator. Instead, there’s probably a polynomial you can expand.
For example, here’s a problem a student asked us via Twitter:

$$\lim_{h \to 0}\dfrac{(h-1)^3 + 1}{h} = ? $$
Early in the semester there’s no way around it: you have to expand the cubic.

Don't remember the formula for expanding a cubic?
Most people don’t. So go with what you do know, and pull out the quadratic, expand it, and then do a few lines of simple algebra.

Begin algebra sub-problem:
(h-1)^3 &= (h-1)(h-1)^2 \\ \\
&= (h-1)(h^2 -2h + 1) \\ \\
&= h(h^2 -2h + 1) – (h^2 -2h + 1) \\ \\
&= (h^3 -2h^2 + h) – (h^2 – 2h + 1) \\ \\
&= h^3 -3h^2 + 3h -1
End algebra sub-problem.

Especially on an exam, don’t let something like this trip you up. Just dive in and do the algebra.

$$(h-1)^3 = h^3 -3h^2 + 3h -1$$
\lim_{h \to 0}\dfrac{(h-1)^3 + 1}{h} &= \lim_{h \to 0}\dfrac{(h^3 -3h^2 + 3h -1) + 1}{h} \\ \\
&= \lim_{h \to 0}\dfrac{h^3 -3h^2 + 3h}{h} \\ \\
&= \lim_{h \to 0}\dfrac{h(h^2 -3h + 3)}{h} \\ \\
&= \lim_{h \to 0}\dfrac{\cancel{h}(h^2 -3h + 3)}{\cancel{h}} \\ \\
&= \lim_{h \to 0} [h^2 – 3h + 3] \\ \\
&= 3 \quad \cmark
We guarantee that if you expand the polynomial in the numerator, and then do a little algebra, the problematic “0” term in the denominator will cancel like it did here. At that point, you can just plug in your value $h = 0$, and be done.
Open to see a different example problem that has a quadratic you need to expand.

\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} &= \text{ ?} \\ \\
&= \lim_{h \to 0}\dfrac{(h^2 -10h + 25) – 25}{h} \\ \\
&= \lim_{h \to 0}\dfrac{h^2 – 10h}{h} \\ \\
&= \lim_{h \to 0}\dfrac{h(h – 10)}{h} \\ \\
&= \lim_{h \to 0}\dfrac{\cancel{h}(h-10)}{\cancel{h}} \\ \\
&= \lim_{h \to 0}[h – 10] \\ \\
&= -10 \quad \cmark

The upshot: Expand the polynomial.
(For now. Later in the course we’ll see how these $\lim_{h \to 0}$ problems relate directly to the definition of the derivative, at which point you can use a shortcut—but not yet. You must know how to solve these using the approach here first. . . and really, it’s just algebra.)

III. Rationalize

Does your problem have some square roots in it, like this?
$$\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} = ?$$
Then rationalize the expression just like you practiced in algebra: multiply both the numerator and denominator by the conjugate $\sqrt{x+5} + \sqrt{5}$.
\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}} \\ \\
&= \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}]} \\ \\
&= \lim_{x \to 0}\dfrac{(x+5) – 5}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\
&= \lim_{x \to 0}\dfrac{x}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\
&= \lim_{x \to 0}\dfrac{\cancel{x}}{\cancel{x}[\sqrt{x+5} + \sqrt{5}]} \\ \\
&= \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} \\ \\
&= \dfrac{1}{2\sqrt{5}} \quad \cmark
We guarantee that if you rationalize the expression, the problematic “0” term in the denominator will cancel like it did here. At that point, you can just plug in for x, and be done.

By the way, notice that in going from the first line above to the second, we multiplied out the terms in the numerators to get rid of the radicals. But we didn’t multiply out the terms in the denominator, because the original x in the denominator cancelled nicely a few steps later.

The upshot: If you have radicals, rationalize.

Of course you need to practice.

Of course reading through our solutions isn’t enough. Instead, you need to practice—and make some mistakes for yourself—so that this is all routine for you when you take your exam. We have lots of problems for you to try, all with complete solutions a single click away so you can quickly check your work (or get unstuck) with no hassle.

For now, we invite you to let us know on the Forum:

  • What tips do you have to share about solving limits problems?
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As of September 2022, we're using our Forum for comments and discussion of this topic, and for any math questions. We'd love to see you there and help! Please tap to visit our Forum:

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3 years ago

I need to find the limit of a cubic function.
“Let lim x->4 f(x)=9 and lim x->4 g(x)= 27. Use the limit rules to find the limit:
26. Find lim x->4 3square root over g(x).

My first thought was getting rid of the quare root so that I have g(x)^2/3. Please help

Reply to  Laura
3 years ago

We’re happy to try to help. We know it’s hard to write math in comments, so first, we assume the question asks you to find
\[\lim_{x \to 4} \sqrt[3]{g(x)} = ?\]
If that’s the question, then limit rules let you move the limit inside the cube-root sign:
\[\lim_{x \to 4} \sqrt[3]{g(x)} = \sqrt[3]{\lim_{x \to 4} g(x)} =\sqrt[3]{27} = 3 \quad \cmark\]
(You probably need to reload your web page to make the math render correctly.)
Hope that does it! And if we didn’t read the question correctly, please let us know. : )

Update September 2022

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