An elementary doubt on computing the limit $ \lim \limits_{x\to0} (\sin x)^x $ Well the question is to compute :$$ \lim \limits_{x\to0} (\sin x)^x $$
Set $y = (\sin x)^x$ Taking logarithm of both sides, we have $ \ln y = x \ln (\sin x) $. Applying limits we would get $$ \lim \limits_{x\to0} \ln y =\lim \limits_{x\to0} x \ln (\sin x)$$ $$ \Rightarrow \ln y = \lim \limits_{x\to0} x \times  \lim \limits_{x\to0} \ln (\sin x) $$ $$ \Rightarrow  \ln y = 0 \times \infty =0$$
$$ \Rightarrow  y = e^0  =1$$
Is my deduction correct here?
Or should (must) I manipulate things to $\lim \limits_{x\to0} x \ln (\sin x) = \lim \limits_{x\to0} \frac{ \ln (\sin x)}{\frac{1}{x}}$ and then apply l'Hôpital's rule and then try and put the limits?
Please explain your answer.
 A: First, a relatively minor point. When $x$ is negative and close to $0$, $\sin x$ is negative, and $(\sin x)^x$ is not defined. Thus, to be precise, we want to calculate 
$$\lim_{x\to 0^+} (\sin x)^x,$$
the limit as $x$ approaches $0$ from the right.
Second, and more importantly, the assertion $0 \times \infty=0$ is, at the very least, highly suspect. Maybe the following example is too simple, but I hope that it will illustrate the issue. Look at the problem of finding 
$$\lim_{x\to 0^+} \;(e^{-1/x})^x.$$
This problem is easy, since for $x\ne 0$, $(e^{-1/x})^x=e^{-1}$, so the limit is $e^{-1}$. But let us try to solve the problem in imitation of your proposal. 
Let $y=(e^{-1/x})^x$. Then $\ln y= x \ln(e^{-1/x})$. Now we deliberately do not notice that $\ln(e^{-1/x})=-1/x$.  Instead, write in imitation of what you did that $\lim_{x\to 0^+}\:x=0$, $\lim_{x\to 0^+} \ln(e^{-1/x})=-\infty$ and "therefore"
$\lim_{x\to 0^+} \ln y=0\times(-\infty)=0$. From this we would conclude that $\lim_{x\to 0^+} \;y =e^0=1$, which is false.
A procedure that sometimes gives the wrong answer is not one that we want to use!
Your problem: There are at least three incorrect things about the $0 \times \infty =0$ argument. One of them has been already discussed in detail. Another is that in fact as $x$ approaches $0$ from the right, $\ln(\sin x)$ becomes large negative.  The third is that if we decide (with proper justification or not) that $\lim_{x\to 0^+}\; \ln y =0$, the conclusion should be that $\lim_{x\to 0^+} y=e^0=1$.
By way of contrast, the L'Hospital's Rule calculation that you propose works well. We have
$$\ln y= \frac{\ln(\sin x)}{\frac{1}{x}}.$$
As $x$ approaches $0$ from the right, the top "approaches" $-\infty$, and the bottom "approaches" $\infty$. We conclude that the limit is equal to 
$$\lim_{x\to 0^+}\frac{\frac{\cos x}{\sin x}}{\frac{-1}{x^2}},$$
if that limit exists. The expression we want to find the limit of can be rewritten as $-\frac{x^2\cos x}{\sin x}$. The limit is easy to calculate, using the fact that $\frac{x}{\sin x}$ approaches $1$. So we conclude that 
$$\lim_{x \to 0^+} y=0$$
and therefore
$$\lim_{x\to 0^+} (\sin x)^x =1.$$ 
Note that apart from a little slip at the end, your $0 \times \infty=0$ procedure gave the right answer. However, this was (sort of) accidental. The procedure is theoretically deeply flawed, and one can give many examples, not just the $(e^{-1/x})^x$ example that was discussed in detail, where the procedure gives the wrong answer.   
A: Edited:
You can never take limits by parts if one part is 0, and the other goes to infinity.
$$
\lim_{x \rightarrow 0} x \ln(\sin x) = 
\lim_{x \rightarrow 0} \sin (x) \ln(\sin x) \times \frac{x}{\sin x}
$$
This is equal to (by putting $\sin x = t$),
$$
\lim_{x \rightarrow 0} \frac{x}{\sin x} \times \lim_{t\rightarrow 0} (t \ln t)
$$
assuming that both limits exists and are not of type $0 \times \infty$ or similar.
Now, the first limit is clearly 1, (standard limit), and the second one is 0,
after a change of variables, $t \mapsto 1/s.$
Thus, the entire limit is 0.
