How to get the following limit into indeterminate form? I am struggling to get the following limit into its indeterminate form so that i can apply the l'Hopitals rule:
$$\lim_{x\to 0^+}(\sin x)^x$$
A solution would be greatly appreciated, been struggling on this one for hours now
 A: $(\sin x)^x \to x \log \sin x = \frac{\log \sin x}{\frac{1}{x}}$
A: As a similar and useful hint, one can use the definition of Infinitesimal Functions here. In fact, while $x\to 0$ we have $\sin x\sim x$ . Now apply the method you know. I mean put $y=x^x$ and then...
A: Try setting the equation equal to y then applying the natural log to each side. I prefer working with ln rather than log because it's easier to derive.
A: $$\begin{align}
\lim_{x\rightarrow 0} (\sin x)^x & = \exp\left(\log \left(\lim_{x\rightarrow 0} (\sin x)^x\right)\right)  \\
& = \exp\left(\lim_{x \rightarrow 0} \log \left( (\sin x)^x\right)\right)\\
&= \exp\left(\lim_{x \rightarrow 0} x \log  (\sin x)\right)\\
&= \exp\left(\lim_{x \rightarrow 0} \dfrac{\log  (\sin x)}{\frac1x}\right)\\
\text{or}&= \exp\left(\lim_{x \rightarrow 0} \dfrac{x}{\frac{1}{\log  (\sin x)}}\right)\\
\end{align}$$
A: Write:
$$e^{x\ln \sin x}=e^{x\ln \frac{\sin x}{x}+x\ln x}$$
If $x\rightarrow 0$ we have
$$\ln \frac{\sin x}{x}\rightarrow 0$$
and
$$e^{x\ln \frac{\sin x}{x}+x\ln x}\sim e^{x\ln x}$$
but
$$\lim_{x\rightarrow 0}\frac{\ln x}{1/x}=\lim_{t\rightarrow +\infty}\frac{-\ln t}{t}=0$$
Then, result is $1$.
