calculus - Indeterminate form $0^0$ using L'hospitals rule when calculating $lim_{xto0^+} x^{sin(x)}$

Given the question $$\lim_{x\to0^+} x^{\sin(x)}$$

I have deducted so far that this has the indeterminate form $0^0$ so I have taken the natural logarithm of both sides to give me: $$\lim_{x\to0^+} \ln(y) = \lim_{x\to0^+}\sin(x)\ln(x)$$ This still has an indeterminate form so I rearrange it to start using L'hospitals rule: $$\lim_{x\to0^+}\frac{\ln(x)}{\csc(x)}$$ $$=\lim_{x\to0^+}\frac{\frac1x}{-\csc(x)\cot(x)}$$ $$=\lim_{x\to0^+}\frac{-\sin(x)\tan(x)}{x}$$ $$=\lim_{x\to0^+}\frac{-\cos(x)\sec^2(x)}{1} = -1$$ $$\lim_{x\to0^+}x^{\sin(x)} = \frac1e$$

I don't think this is correct, am I doing something wrong, and if so, where?

Answer

$$\lim_{x\to0^+} x^{\sin(x)}$$

$$=\lim_{x\to0^+} \exp(\ln(x)\sin(x))$$

$$=\exp(\lim_{x\to0^+} \ln(x)\sin(x))$$

$$=\exp(\lim_{x\to0^+} \frac{\ln(x)}{1/\sin(x)})$$

Applying L'Hôpital's rule:

$$=\exp(\lim_{x\to0^+} -\frac{\sin(x)^2}{x\cos(x)})$$

$$=\exp(-1\cdot \lim_{x\to0^+} \frac{1}{\cos(x)}\lim_{x\to0^+} \frac{\sin(x)^2}{x})$$

$$=\exp(-\lim_{x\to0^+} \frac{\sin(x)^2}{x})$$

Applying L'Hôpital's rule:

$$=\exp(-\lim_{x\to0^+} 2\cos(x)\sin(x))$$

$$\color{grey}{e^0=1}$$

$$=\boxed {\color{blue}1}$$