The limit $\left( \sin x \right)^{x}$ when $x \rightarrow\; 0$ Take the limit $\left( \sin x \right)^{x}$ when $x \rightarrow\; 0$
I tried using $e^{\ln \left( \left( \sin x \right)^{x} \right)}=e^{x\cdot \ln \left( \sin x \right)}$ and then saying sinx → 0 but ln(0) is undefined. Stopped there.
 A: Hint: What happens  with $\left(\sin n\pi\right)^{n\pi}$ and with
$\left(\sin\left(\frac{1}{2}\pi+2n\pi\right)\right)^{\frac{1}{2}\pi+2n\pi}$ if you look at integers $n\rightarrow\infty$?
Edit
My answer is not relevant anymore. It deals with $x\rightarrow\infty$ wich was the original question. Later that was changed into $x\rightarrow 0$ by an edit.
A: For $x$ near zero, $\sin x\sim x$, so you are looking at $x\log x$ when $x\to0$, which goes to zero. This suggests the limit is $e^0=1$. 
We can confirm this by using L'Hopital:
$$
x\log\sin x=\frac{\log\sin x}{1/x},
$$
which is of the form "$\infty/\infty$". So
$$
\lim_{x\to0}x\log\sin x=\lim_{x\to0}\frac{\log\sin x}{1/x}=\lim_{x\to0}\frac{\cos x}{\sin x}\,\frac{1}{\frac{-1}{x^2}}=\lim_{x\to0}\frac{-x\cos x}{\frac{\sin x}x}=0
$$
Edit: here is a way to see that the sine function has no impact in how this limit behaves: it involves the change of variable $t=\sin x$. 
$$
\lim_{x\to0}x\log\sin x=\lim_{x\to0} \frac{x}{\sin x}\,\sin x\,\log\sin x=\lim_{x\to0} \frac{x}{\sin x}\,\lim_{x\to0} \sin x\log \sin x\\ =\lim_{x\to0} \sin x\log x=\lim_{t\to0}\,t\log t=0
$$
Edit 2: for the sake of completeness, I will include a proof of $\lim_{x\to0}x\log x=0$ without using L'Hopital. All we need is the Mean Value Theorem and some basic knowledge about the exponential and logarithm functions. 
First, note that
$$
\lim_{x\to0}x\log x=\lim_{x\to0^+}x\log x=\lim_{u\to\infty}\frac1u\,\log\frac1u=-\lim_{u\to\infty}\frac1u\,\log u=-\lim_{t\to\infty}\frac{t}{e^t}.
$$
To calculate this last limit, we will use the MVT. Given $s>0$, there exists $\xi\in(0,s)$ with 
$$
e^s-1=e^{\xi}(s-0)=se^\xi.
$$
As $e^\xi\geq1$, we get $e^s-1\geq s$, or $e^s\geq1+s$. Integrating from $0$ to $t$, we get (as both sides of the inequality are positive functions, integrating will preserve the inequality)
$$
e^t-1=\int_0^te^s\,ds\geq\int_0^t(1+s)\,ds=t+\frac{t^2}2.
$$
Then $e^t\geq1+t+\frac{t^2}2\geq\frac{t^2}2$ (of course, if you know Taylor's expansion, you can get this right away). Now, for $t>0$,
$$
0\leq\frac{t}{e^t}\leq\frac{2t}{t^2}=\frac2t.
$$
As $2/t\to0$ as $t\to\infty$, we get $\lim_{t\to\infty} t/e^t=0$.
