Early in your Calculus studies, there are two “special trig limits” that you simply have to memorize.

We’ll be able prove them easily later, but for now please commit to memory:

\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*}

You’ve seen the Special Limit I before, when we looked at “Some Limits that Do Exist; Some that Do Not.” The left-hand figure might remind you of your exploration there. The right-hand figure shows the second special limit.

Typically you’ll have to do a little “massaging” of the expression you’ve been given in order to be able to use the special trig limits, as the following example illustrates.

Special Trig Limit Example 1: $\displaystyle{\lim_{x \to 0}\dfrac{\sin(5x)}{x}}$

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

*Solution*.

The expression in the question reminds us of the first “Special Trig Limit,”

\[\lim_{x \to 0}\frac{\sin(x)}{x} = 1 \]
But it isn’t quite the same, because in our expression the argument of sin that’s in the numerator (5*x*) doesn’t match what’s in the denominator (*x*). That is, since we have $\sin(5x)$ in the numerator, we need $5x$ in the denominator.

So let’s multiply the expression by $\dfrac{5}{5}$, and then do some rearranging:

\begin{align*}

\lim_{x \to 0}\dfrac{\sin(5x)}{x} &= \lim_{x \to 0} \dfrac{\sin(5x)}{x} \cdot \frac{5}{5} \\[8px]
&= 5 \cdot \lim_{x \to 0} \frac{\sin(5x)}{5x} &\text{[Recall $\displaystyle{\lim_{x \to

0}\dfrac{\sin(\text{whatever})}{(\text{the same whatever})} =1 }$]}\\[8px]
&= 5 \cdot 1 = 5 \quad \cmark

\end{align*}

The following problems will let you practice using similar techniques.

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This next problem doesn’t make use of the special trig limits, but instead shows how you may need to do some other trig-manipulation in order to compute the limit.

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- Early in the semester there are two “special trig limits” you simply must remember:

\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*} - Also remember your trig identities and such, which you may also need to use.

The tactics we’ve introduced in this Section, and that you’ve practiced, will let you compute almost every limit at a point that you’ll be asked to find early in your Calculus studies. In the next Section we’ll investigate limits as a function goes “to infinity”!

Questions or comments about the material on this screen, or anything regarding limits? Please post on the Forum!