Question

L=^{sin x}" alt="\lim_{x \to \0} x^{sin x}" align="absmiddle" class="latex-formula">

Answer 1

By applying the exponential and logarithmic functions, we have

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

Then in the limit,

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \lim_{x\to0} \exp \left(\ln \left(x^{\sin(x)}\right)\right)$

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \ln \left(x^{\sin(x)}\right)\right)$

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \sin(x) \ln(x) \right)$

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \frac{\ln(x)}{\csc(x)} \right)$

As x approaches 0 (from the right), both ln(x) and csc(x) approach infinity (ignoring sign). Applying L'Hopitâl's rule gives

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \frac{\frac1x}{-\csc(x)\cot(x)} \right) = \exp \left( -\lim_{x\to0} \frac{\sin^2(x)}{x\cos(x)} \right)$

Recall that

$\displaystyle \lim_{x\to0} \frac{\sin(x)}{x} = 1$

Then

$\displaystyle \lim_{x\to0} \frac{\sin^2(x)}{x\cos(x)} = \lim_{x\to0} \frac{\sin(x)}{\cos(x)} = \lim_{x\to0} \tan(x) = 0$

So, our limit is

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp(0) = \boxed{1}$