# Limit x tends to 0+ sinx^sinx ?

Jul 12, 2016

The answer is ${\lim}_{x \to {0}^{+}} {\left(\sin \left(x\right)\right)}^{\sin \left(x\right)} = 1$.

#### Explanation:

First, let $y = {\left(\sin \left(x\right)\right)}^{\sin \left(x\right)}$. Then $\ln \left(y\right) = \sin \left(x\right) \ln \left(\sin \left(x\right)\right) = \frac{\ln \left(\sin \left(x\right)\right)}{\csc} \left(x\right)$. Now use L'Hopital's Rule to evaluate the limit of this expression (it is an $\frac{\infty}{\infty}$ indeterminate form).

${\lim}_{x \to 0 +} \frac{\ln \left(\sin \left(x\right)\right)}{\csc} \left(x\right) = {\lim}_{x \to 0 +} \frac{\frac{1}{\sin} \left(x\right) \cdot \cos \left(x\right)}{- \cot \left(x\right) \cdot \csc \left(x\right)}$

$= {\lim}_{x \to 0 +} \left(- \tan \left(x\right)\right) = 0$

Therefore, $\ln \left({\lim}_{x \to 0 +} y\right) = {\lim}_{x \to 0} \ln \left(y\right) = {\lim}_{x \to 0 +} \sin \left(x\right) \ln \left(\sin \left(x\right)\right) = 0$. Exponentiation now implies that

${\lim}_{x \to 0 +} {\left(\sin \left(x\right)\right)}^{\sin \left(x\right)} = {\lim}_{x \to 0 +} y = {e}^{0} = 1$.

The graph of ${\left(\sin \left(x\right)\right)}^{\sin \left(x\right)}$ confirms this visually.

graph{sin(x)^(sin(x)) [-4.054, 4.065, -2.034, 2.025]}