what is the limit? $$\lim_{x\to0+}  (\sin x)^{2/\ln x}$$
Not too sure what method to use. I have tried to follow all the helpful answers and this is my working I have done. Please tell me if I have made a mistake. Many thanks! 

is this answer correct?
 A: $$\lim_{x\to 0^+} x\cot x = \lim_{x\to 0^+} \frac{x}{\sin x}\cdot\cos x = 1, $$
hence by De l'Hopital theorem:
$$ \lim_{x\to 0^+}\frac{\log\sin x}{\log x} = \lim_{x\to 0^+} x\cot x = 1, $$
and by exponentiating, then squaring, both sides:
$$ \lim_{x\to 0^+}\left(\sin x\right)^{\frac{2}{\log x}} = \color{red}{e^2}.$$
A: Let $$\displaystyle y = \lim_{x\rightarrow 0^{+}}(\sin x)^{\frac{2}{\ln x}}\;,$$ Now put $x=0+h\;,$ Then  $h\rightarrow 0\;,$ Where $h$ is a small positive quantity.
So $$\displaystyle y=\lim_{h\rightarrow 0}(\sin h)^{\frac{2}{\ln h}}\;,$$ Now taking $\ln $ on both side, We get
$$\displaystyle \ln(y) = \lim_{h\rightarrow 0}\frac{2\cdot \ln (\sin h)}{\ln h}\; \left(\frac{\infty}{\infty}\right)$$ form
So we will Use $\bf{D\; L\; Hopital\; Rule}$
So $$\displaystyle \ln(y) = \lim_{h\rightarrow 0} \frac{2\cdot \cos h\cdot h}{\sin h} = 2\Rightarrow \ln(y) = 2\Rightarrow y=e^{2}$$
A: The limit is 0, you put the value in place of x. If the answer results in any real number divided by 0 then it is undefined equation and you'll have to simplify it first before putting the value of limit.
Here you can try to apply the limit directly and if it results in 0/0 form then its not correct, simplify it first and then apply the limit.
A: A different approach is to make the substitution $x\to e^x$. Then, 
$$\begin{align}
\lim_{x\to 0^+}\left(\sin x\right)^{2/\log x}=\lim_{x\to -\infty}\left(\sin (e^x)\right)^{2/x}
\end{align}$$
Since $$\sin (e^x)=e^x+O(e^{3x})$$
we have $\sin (e^x)^{2/x}=e^2\left(1+O(e^{2x})\right)^{2/x}=e^2\left(1+\frac{2}{x}O(e^{2x})\right)\to e^2\,\,\text{as}\,\,x\to -\infty$
Therefore, 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0^+}\left(\sin x\right)^{2/\log x}=e^2}$$
A: Let $f(x) = (\sin x)^{2/\log x}$ and we need to evaluate $\lim_{x \to 0^{+}}f(x)$. If this limit is $L$ then
\begin{align}
\log L &= \log\left(\lim_{x \to 0^{+}}f(x)\right)\notag\\
&= \lim_{x \to 0^{+}}\log f(x)\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0^{+}}\frac{2\log \sin x}{\log x}\notag\\
&= 2\lim_{x \to 0^{+}}\dfrac{\log \left(\dfrac{\sin x}{x}\right) + \log x}{\log x}\notag\\
&= 2\left(\lim_{x \to 0^{+}}\dfrac{\log \left(\dfrac{\sin x}{x}\right)}{\log x} + 1\right)\notag\\
&= 2(0 + 1) = 2\notag
\end{align}
Hence $L = e^{2}$. As with most of the limit problems, LHR / Taylor is not required here.
