## 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.

This is the first thing you should always try: just plug the value of

*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.

Find $\displaystyle{ \lim_{x \to -2}\left(x^3 +4\right) }$.

Find $\displaystyle{ \lim_{x \to 1} }\dfrac{\ln x}{x}$.

Find $\displaystyle{\lim_{x \to 1} \dfrac{3}{x-1} }$.

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):

\[\text{If} \quad \lim_{x \to a}f(x) = \frac{\text{a non-zero number}}{0}\]

\[\text{then the limit does not exist (DNE).}\]

Frequently (on homework and exams, anyway) when you try Substitution, you obtain the fraction $\dfrac{0}{0}$, 0 divided by 0. When this happens, if you can factor the numerator and/or the denominator, do so. A common term will cancel, removing the problematic $0$ in the denominator. Guaranteed.

**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.

$\displaystyle{\lim_{x \to 3} \frac{x-3}{x^2-9}}$

Find $\displaystyle{ \lim_{x \to 1} }\frac{x^2 + 4x -5}{x^2 – 1}$.

If you try Substitution and obtain the fraction $\dfrac{0}{0}$ (0 divided by 0) and the expression has a square root in it, then rationalize the expression just like you practiced in Algebra. That is, multiply the numerator and the denominator by the conjugate of the part that has the square root. The following example illustrates.

**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.

Find $\displaystyle{\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}}$.

Find $\displaystyle{ \lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3}.}$

For some problems, when you try Substitution and obtain $\dfrac{0}{0}$, you just use your algebraic skills to transform the expression into something where Substitution will work. For instance, if there’s a quadratic, expand it. Or if you have some fractions in the numerator, put them over a common denominator.

Find $\displaystyle{\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h}}$.

When your problem has a trig function in it (sin, cos, tan, …), often you need to do some manipulation in order to find the limit. To help:

1. Remember your fundamental trig identities. For instance:

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

Our Handy Table of Trig Formulas and Identities could be useful. It’s always available via the “Reference” link at the very top of every page.

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.

Find $\displaystyle{\lim_{x \to 0}\dfrac{1- \cos(x) }{\sin^2(x)} }$.

Find $\displaystyle{\lim_{x \to 0}\dfrac{\sin(5x)}{x} }$.

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