# How do you find the limit of x^(sin(x)) as x approaches 0?

Sep 27, 2016

$1$

#### Explanation:

let $L = {\lim}_{x \to 0} {x}^{\sin x}$

$\implies \ln L = \ln {\lim}_{x \to 0} {x}^{\sin x}$

$= {\lim}_{x \to 0} \ln {x}^{\sin x}$

$= {\lim}_{x \to 0} \sin x \ln x$

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

$= {\lim}_{x \to 0} \frac{\ln x}{\csc x}$

this is in indeterminate $\frac{\infty}{\infty}$ form so we can use L'Hôpital's Rule

$= {\lim}_{x \to 0} \frac{\frac{1}{x}}{- \csc x \cot x}$

$= - {\lim}_{x \to 0} \frac{\sin x \tan x}{x}$

Next bit is unnecessary, see ratnaker-m's note below...

this is now in indeterminate $\frac{0}{0}$ form so we can go again

$\ln L = - {\lim}_{x \to 0} \frac{\cos x \tan x + \sin x {\sec}^{2} x}{1}$

$= - 0$

So:
$L = {e}^{- 0} = 1$

Sep 27, 2016

$1$

#### Explanation:

x^sin x = (1+x-1)^sinx = 1 +sin x/(1!)(x-1)+(sinx(sinx-1))/(2!)(x-1)^2+cdots+

then

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