## 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:**We now have

*much*more interactive ways for you to learn about the foundational concept of Limits, making heavy use of interactive Desmos graphing calculators so you can work with these ideas for yourself, and develop your problem solving skills step-by-step. Please visit our Limits Chapter to

*really*get this material down for yourself.

It’s all free, and waiting for you! (Why? Just because we’re educators who believe you deserve the chance to develop a better understanding of Calculus for yourself, and so we’re aiming to provide that. We hope you’ll take advantage!)

### I. Factoring

$$\lim_{x \to 4}\dfrac{x^2 – 16}{x-4} = ?$$

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}

$$

**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:

\begin{align*}

\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

\end{align*}

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.

**The upshot:**Factor whenever you can.

**RELATED MATERIAL**

- 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.

Then

\begin{align*}

\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

\end{align*}

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.

**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 $\sqrt{x+5} + \sqrt{5}$.

\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}} \\ \\

&= \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

\end{align*}

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.

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*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:

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### What are your thoughts or questions?

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

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. : )

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