# Special Limits Involving sin(x), x, and tan(x)

## Key Questions

• We will use l'Hôpital's Rule.

l'Hôpital's rule states:

${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

In this example, $f \left(x\right)$ would be $\sin x$, and $g \left(x\right)$ would be $x$.

Thus,

${\lim}_{x \to 0} \frac{\sin x}{x} = {\lim}_{x \to 0} \frac{\cos x}{1}$

Quite clearly, this limit evaluates to $1$, since $\cos 0$ is equal to $1$.

By using: ${\lim}_{x \to 0} \frac{\sin x}{x} = 1$, ${\lim}_{x \to 0} \frac{\tan x}{x} = 1$.

Explanation
Let us look at some details.
Since $\tan x = \frac{\sin x}{\cos x}$,

${\lim}_{x \to 0} \frac{\tan x}{x} = {\lim}_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x}$

by the Product Rule,

$= \left({\lim}_{x \to 0} \frac{\sin x}{x}\right) \cdot \left({\lim}_{x \to 0} \frac{1}{\cos x}\right)$

by ${\lim}_{x \to 0} \frac{\sin x}{x} = 1$,

$= 1 \cdot \frac{1}{\cos \left(0\right)} = 1$

• One of some useful limits involving $y = \sin x$ is
${\lim}_{x \to 0} \frac{\sin x}{x} = 1$.
(Note: This limit indicates that two functions $y = \sin x$ and $y = x$ are very similar when $x$ is near $0$.)