Evaluating $\lim_{x\to0}(x\tan x)^x$ Any suggestions for evaluating the limit $$\lim_{x\to0}(x\tan x)^x$$
I have tried writing $\tan$ as $\dfrac{\sin}{\cos}$ and then got the Taylor series of them but it didn't lead me somewhere. Thanks a lot
 A: Assuming the limit exists and is equal to $L$, take logs:
$$\log{L} = \lim_{x \to 0} x \log{(x \tan{x})}$$
Use $\tan{x} \sim x$ in this limit.  Then use 
$$\lim_{y \to 0} y \log{y} = 0$$
and the limit should come out easily:
$$\log{L} = \lim_{x \to 0} x \log{x^2} = \lim_{x \to 0} 2 x \log{x} = 0$$
Therefore, $L=1$.
A: $$\large(x\tan x)^x=e^{\log(x\tan x)^x}=e^{x\log(x\tan x)}$$
Then $$\large\lim_{x\to 0}(x\tan x)^x=e^{\lim_{x\to 0}x\log(x\tan x)}\tag{since $\exp(x)$ is  continuous}$$
Now $$\lim_{x\to 0}x\log(x\tan x)=\lim_{x\to 0}\dfrac{\log x+\log\tan x}{\dfrac{1}{x}}$$
Apply L'Hopital's rule.
A: The limit exists if and only if
$$
\lim_{x\to0} \log((x\tan x)^x)=\lim_{x\to0}(x\log(x\sin x)-x\log\cos x)
$$
exists, since the exponential is a continuous (increasing) function. The second summand gives no problem, because its limit is clearly $0$. So we compute
\begin{align}
\lim_{x\to0}x\log(x\sin x)&=
\lim_{x\to0}x\log\left(x^2\frac{\sin x}{x}\right)\\
&=\lim_{x\to0}2x\log x +\lim_{x\to0}x\log\left(\frac{\sin x}{x}\right).
\end{align}
Again, the second summand has limit $0$ and also the first one is well-known to have limit $0$. Therefore
$$
\lim_{x\to0} \log((x\tan x)^x)=0
$$
and so
$$
\lim_{x\to0} (x\tan x)^x)=e^0=1.
$$
A: Hint:let $y=(x\tan x)^x$ , use  $ x\sim \tan x$ and $\lim_{x\to0}x\ln x=0$ $$\large{\lim_{x\to0}y=\lim_{x\to0}e^{\ln y}=\lim_{x\to0}e^{x\ln(x\tan x)}=\lim_{x\to0}e^{x\ln(x^2)}=\lim_{x\to0}e^{2x\ln(x)}=e^{0}=1}$$
