# How do you find lim_(x->0^+)(sinx)^x?

## How do you find ${\lim}_{x \to {0}^{+}} {\left(\sin x\right)}^{x}$?

Jan 18, 2018

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

#### Explanation:

lim_(xrarr0^+)sinx^x=?

$\sin {x}^{x} = {e}^{\ln {\left(\sin x\right)}^{x}} = {e}^{x \ln \left(\sin x\right)}$

$= {\lim}_{x \rightarrow {0}^{+}} {e}^{x \ln \left(\sin x\right)}$ $\left(1\right)$

You see that we still have problem and can't calculate the limit at it's current form as $\ln 0$ is not defined.
Let's study this limit seperately

${\lim}_{x \rightarrow {0}^{+}} \ln \left(\sin x\right) =$

Set

$\sin x = u$
$x \to {0}^{+}$
$u \to {0}^{+}$

$=$ ${\lim}_{u \rightarrow {0}^{+}} \ln u = - \infty$

graph{lnx [-10, 10, -5, 5]}

${\lim}_{x \rightarrow {0}^{+}} x \ln \left(\sin x\right) = {\lim}_{x \rightarrow {0}^{+}} \ln \frac{\sin x}{\frac{1}{x}} {=}_{D L H}^{\left(\frac{- \infty}{+ \infty}\right)}$

${\lim}_{x \rightarrow {0}^{+}} \frac{\frac{\left(\sin x\right) '}{\sin} x}{- \frac{1}{x} ^ 2} = - {\lim}_{x \rightarrow {0}^{+}} \frac{\cos \frac{x}{\sin} x}{\frac{1}{x} ^ 2} = - {\lim}_{x \rightarrow {0}^{+}} \left(\frac{{x}^{2} \cos x}{\sin} x\right)$

$x \to {0}^{+}$

$x > 0$

$=$ -lim_(xrarr0^+)(cosx)/(sinx/x*1/x)=^((1/(1*(+oo))) $0$

• because

${\lim}_{x \rightarrow {0}^{+}} \cos x = \cos 0 = 1$

${\lim}_{x \rightarrow {0}^{+}} \sin \frac{x}{x} = 1$

lim_(xrarr0^+)(1/x)=^((1/(0^+)) $+ \infty$

Now if in $\left(1\right)$ i set:

$y = x \ln \left(\sin x\right)$

$x \to {0}^{+}$
$y \to 0$

and $\left(1\right)$ now becomes:

${\lim}_{y \rightarrow 0} {e}^{y} = {e}^{0} = 1$

As a result,

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