By applying the exponential and logarithmic functions, we have
[tex]x^{\sin(x)} = \exp \left(\ln \left(x^{\sin(x)}\right)\right)[/tex]
Then in the limit,
[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \lim_{x\to0} \exp \left(\ln \left(x^{\sin(x)}\right)\right)[/tex]
[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \ln \left(x^{\sin(x)}\right)\right)[/tex]
[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \sin(x) \ln(x) \right)[/tex]
[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \frac{\ln(x)}{\csc(x)} \right)[/tex]
As x approaches 0 (from the right), both ln(x) and csc(x) approach infinity (ignoring sign). Applying L'Hopitâl's rule gives
[tex]\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)[/tex]
Recall that
[tex]\displaystyle \lim_{x\to0} \frac{\sin(x)}{x} = 1[/tex]
Then
[tex]\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[/tex]
So, our limit is
[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp(0) = \boxed{1}[/tex]
