Limit $\lim_{x\to 0^{+}} (\tan x)^x$ $$\lim_{x\to 0^{+}} (\tan x)^x$$
$$\lim_{x\to 0^{+}} e^{\ln((\tan x)^x)}=\lim_{x\to 0^{+}} e^{x\ln(\tan x)}=\lim_{x\to 0^{+}} e^{x[\ln(\sin x)-\ln(\cos x)]}$$
We can continue to create an expression that may help us use L'Hospital but it does not seem to be correct 
P.S can we write:
$$1=(\frac{-1}{-1})^x\leq \lim_{x\to 0^+}\Bigl(\frac{\sin x}{\cos x}\Bigr)^x\leq \Bigl(\frac{1}{1}\Bigr)^x=1$$?
 A: $\lim_{x \to 0+} x \ln (\tan x)=\lim_{x \to 0+} \frac {\ln (\tan x)} {1/x}=-\lim_{x \to 0+} \frac {\sec^{2}x} {\tan x /x^{2}}$. You can write this as $-\lim_{x \to 0+} \frac {x^{2}} { \sin x \cos x}$ and this limit is $0$. (Why?). Hence the given limit is $e^{0}=1$. 
A: $$\lim_{x\to0^+}(\tan x)^x=\left(\lim_{x\to0^+}\tan x^{\tan x}\right)^{\lim_{x\to^+}\dfrac x{\tan x}}$$
For the inner limit use Limit of $x^x$ as $x$ tends to $0$
A: Compute first the limit of the logarithm, using equivalence:
$\tan x\sim_0 x,\:$ so 
$\quad \ln\bigl((\tan x)^x\bigr)=x\ln(\tan x)\sim_0x\ln x \xrightarrow[x\to 0]{} 0.$
A: You surely know


*

*$\lim_{x\to 0}\frac{\sin x}{x} = 1 \Rightarrow \lim_{x\to 0}\frac{\tan x}{x} = \lim_{x\to 0}\left(\frac{\sin x}{x}\cdot \frac 1{\cos x} \right) = 1$.

*Besides this, it is easy to show that $\lim_{x\to 0^+}x^x = 1$.


Hence,
$$(\tan x)^x = \underbrace{\left(\frac{\tan x}{x}\right)^x}_{\stackrel{x\to 0^+}{\longrightarrow}1^0=1}\cdot \underbrace{x^x}_{\stackrel{x\to 0^+}{\longrightarrow}1} \stackrel{x\to 0^+}{\longrightarrow}1$$
