# How do you evaluate lim_(x-> 0) tanx/x?

Aug 24, 2017

Since $\tan \left(0\right) = \sin \frac{0}{\cos} \left(0\right) = \frac{0}{1} = 0$ and the denomianator equals $0$, we can use L'Hospitals rule .

$L = {\lim}_{x \to 0} \frac{{\sec}^{2} x}{1}$

$L = {\lim}_{x \to 0} {\sec}^{2} x$

$L = {\sec}^{2} \left(0\right)$

$L = \frac{1}{\cos} ^ 2 \left(0\right)$

$L = 1$

If we check the graph of the function $f \left(x\right) = \tan \frac{x}{x}$, we see that when $x \to 0$, $y$ approaches $1$.

Hopefully this helps!

Aug 25, 2017

#### Explanation:

$\tan x = \sin \frac{x}{\cos} x$ so

$\tan \frac{x}{x} = \sin \frac{x}{x \cos x} = \sin \frac{x}{x} \cdot \frac{1}{\cos} x$.

So,

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

$= {\lim}_{x \rightarrow o} \sin \frac{x}{x} \cdot {\lim}_{x \rightarrow o} \frac{1}{\cos} x$

$= 1 \cdot \frac{1}{1} = 1$