How to find $\lim_\limits{x\to0}(x \sin x)^{\tan x}$? How can I find $\lim\limits_{x \to 0} (x\sin x)^{\tan x}$?
My problem is: if we use logarithm to solve this limit we will not only solving to positives $x$? I mean it would not be a limit by the right side? How can I solve this limit by the left side without loss of generality?
I'm sorry for the mistakes in writing and thank you :)
 A: Hint: Let $L=(x\sin x)^{\tan x}$, then $\ln L=\tan x \ln (x\sin x)=\frac{\ln (x\sin x)}{\cot x}$. Now you can apply L' Hospital rule.
A: Note the following standard limit $$\lim_{y \to 0^{+}}y^{a}\log y = 0\tag{1}$$ for $a > 0$ which can be used here. If $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}(x\sin x)^{\tan x}\right\}\notag\\
&= \lim_{x \to 0}\log(x\sin x)^{\tan x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\tan x\log(x\sin x)\notag\\
&= \lim_{x \to 0}\frac{\sin x}{\cos x}\log(x\sin x)\notag\\
&= \lim_{x \to 0}\sin x\log(x\sin x)\notag\\
&= \lim_{x \to 0}\frac{\sin x}{x}\cdot x \cdot\log\left(x^{2}\cdot\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0}x \cdot\log(x^{2}) + x\log\left(\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0}x \cdot\log(x^{2})\notag
\end{align}
From $(1)$ it is easy to see that the last limit is $0$ (put $y = x^{2}$ and $a = 1/2$ and we need to consider $x \to 0^{+}$ and $x \to 0^{-}$ separately). Hence $\log L = 0$ and $L = 1$.
A: This is not a full answer since addressing only the problem of the limit when $x\to 0^+$.
For the calculation of the limit, consider $$A=\big(x \sin (x)\big)^{\tan x}$$ Take logarithms $$\log(A)=\tan(x) \log\big(x \sin (x)\big)=\tan(x)\Big( \log(x)+ \log\big( \sin (x)\big)\Big)$$ Now, Taylor series $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)=x\big(1-\frac{x^2}{6}+O\left(x^4\right)\big)$$ $$\log\big( \sin (x)\big)=\log(x)-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log(x)+\log\big( \sin (x)\big)=2\log(x)-\frac{x^2}{6}+O\left(x^4\right)$$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ So, combining all the pieces, $$\log(A)=2 x \log (x)+x^3 \left(\frac{2 \log (x)}{3}-\frac{1}{6}\right)+O\left(x^4\right)$$ Now, using $A=e^{\log(A)}$ and Taylor again $$A=1+2 x \log (x)+2 x^2 \log ^2(x)+O\left(x^3\right)$$ which shows the limit and how it is approached.
For illustration purposes, let us use $x=0.1$ (which is quite large). The exact value is $\approx 0.629880$ while the approximation gives $\approx 0.645521$.
Probably more impressive will be $x=0.01$.  The exact value is $\approx 0.912008$ while the approximation gives $\approx 0.912138$. 
