Indeterminate powers and limits The question is $\lim\limits_{x\rightarrow 0^+} x^{8 \sin(x)}$. It says, use L'Hospital's rule if necessary. Are there other methods to solve this? L'Hospital's rule would be complicated to evaluate, and might lead to other indeterminate forms. I tried to take the natural log and solve, but I failed. Any other methods I could use?  I would ike a step-by step walk-through to see if it is possible to use L'Hospital's rule?
 A: We know the elementary result in analysis
$$\sin x\sim_0 x\qquad\mathrm{and}\qquad \lim_{x\to 0^+}x\log x=0$$
then we have
$$\lim\limits_{x\rightarrow 0^+}x^{8\sin x}=\lim\limits_{x\rightarrow 0^+}e^{8(\sin x )(\log x)}=\lim\limits_{x\rightarrow 0^+}e^{8x\log x}=e^0=1$$
A: $$\lim\limits_{x\rightarrow 0^+}x^{8\sin x}=\lim\limits_{x\rightarrow 0^+}e^{8(\sin x )(\log x)}$$
Now $$\lim\limits_{x\rightarrow 0^+}(\sin x \log x)\geq \lim\limits_{x\rightarrow 0^+}x\log x=\lim\limits_{x\rightarrow 0^+}\dfrac{\log x}{\frac{1}{x}}=(DL'H)$$
$$=\lim\limits_{x\rightarrow 0^+}\dfrac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim\limits_{x\rightarrow 0^+}(-x)=0$$ 
Also again using De L'Hospital :
$$\lim\limits_{x\rightarrow 0^+}(\sin x \log x)\leq \lim\limits_{x\rightarrow 0^+}(\frac{2}{\pi}x\log x)=0$$ 
Hence: $\lim\limits_{x\rightarrow 0^+}(\sin x \log x)=0
$
Therefore: 
$$\lim\limits_{x\rightarrow 0^+}x^{8\sin x}=\lim\limits_{x\rightarrow 0^+}e^{8(\sin x )(\log x)}=e^{8\cdot 0}=1$$ 
A: HINT:
$$\text{ Let } y=x^{\sin x}\implies \ln y=\frac{\ln x}{\csc  x}$$
Now, $\lim_{x\to0^+}\frac{\ln x}{\csc  x}$ is of the form  $\frac \infty\infty$
So, applying L'Hospital's rule $$\lim_{x\to0^+} \ln y=\lim_{x\to0^+}\frac{\frac1x}{-\csc x\cot x}=\frac{-1}{\lim_{x\to0^+}\cos x}\cdot\left(\lim_{x\to0^+}\frac{\sin x}x\right)^2\cdot \lim_{x\to0^+}x=0$$ as $\lim_{x\to0}\frac{\sin x}x=1$
