2
$\begingroup$

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 ?

$\endgroup$
1
$\begingroup$

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 $e^0=1$.

$\endgroup$
  • 4
    $\begingroup$ Why the derivative of $ x^{-1} $ is $ x^{-2}$ ? Why not $ -x^{-2}$ ? $\endgroup$ – Out Of Bounds Feb 6 '14 at 21:34
0
$\begingroup$

Have you tried using $(sin \ x)^x=e^{log(sin \ x )x}$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.