# Evaluating $\lim_{x\to 0^+} (\sin\ x)^x$

Evaluate the following limit.

$$\lim_{x\to 0^+} (sin\ x)^x$$

What i have tried:

$$ln\ [\lim_{x\to 0^+} (sin\ x)^x]$$

$$\lim_{x\to 0^+} ln\ (sin\ x)^x$$

$$\lim_{x\to 0^+} \frac{ln\ (sin\ x)}{\frac{1}{x}}$$

Applying l'hopital's rule.

$$\lim_{x\to 0^+} \frac{cot\ x}{-x^{-2}}$$

If i keep applying l'hopital's rule, i get indeterminate form. Is what Iam doing right ?

We have $(\sin x)^x=\exp(x\ln\sin x)$, so let's investigate $\lim_{x\to0^+}x\ln\sin x$ first. Your idea to write the expression as $\frac{\ln\sin x}{\frac1x}$ is fine as it allows us to use l'Hopital: $$\lim_{x\to0^+}x\ln\sin x=\lim_{x\to0^+}\frac{\ln\sin x}{\frac1x}=\lim_{x\to 0^+}\frac{\cot x}{x^{-2}}.$$ The trick as often is to rearrange numerator and denominator suitably. Here try $$\tag1 \frac{\cot x}{x^{-2}}=\frac{x^2\cos x}{\sin x}=\frac x{\sin x}\cdot x\cdot \cos x$$ You should know that $\lim_{x\to 0}\frac{\sin x}x=1$, hence the limit of $(1)$ is simply $1\cdot 0\cdot 1=0$, so that the final answer is $e^0=1$.
• Why the derivative of $x^{-1}$ is $x^{-2}$ ? Why not $-x^{-2}$ ? – Out Of Bounds Feb 6 '14 at 21:34
Have you tried using $(sin \ x)^x=e^{log(sin \ x )x}$?