Solving $\lim_{ x \to 0^+} \sqrt{\tan x}^{\sqrt{x}}$ The limit to be calculated is:
$$\lim_{x \to 0^+}\sqrt{\tan x}^{\sqrt{x}}$$
I tried:
$$L =\lim_{x \to 0^+}\sqrt{\tan x}^{\sqrt{x}}$$
$$\log L = \lim _{x\to 0^+} \ \ \dfrac{1}{2}\cdot\dfrac{\sqrt{x}}{\frac{1}{\log(\tan x)}}$$
To apply the L'hospital theorem but failed, as it went more complex.
How can we solve this, with or without L'Hospital?

$$\begin{align}
\left(\sqrt{\tan(x)}\right)^{\sqrt{x}}&=e^{\frac12\sqrt{x}\log(\sin(x))-\frac12\sqrt{x}\log(\cos(x))}\\\\
&\underbrace{e^{-\frac12\sqrt{x}\log(\cos(x))}}_{\to 1}\,\underbrace{e^{\frac12\sqrt{x}\log\left(\frac{\sin(x)}{x}\right)}}_{\to 1}\,\underbrace{e^{\sqrt{x}\log(\sqrt{x})}}_{\to 1}\\\\
\end{align}$$
since we recall that $\lim_{y\to 0}y\log(y)=0$ (i.e., with $y=\sqrt{x}$), which can be shown a variety of ways including application of L'Hospital's Rule.

Note that we could have arrived at this result quickly by using asymptotic analysis.  
Here $\tan(x)\sim x$ as $x\to 0$.  
Then we are evaluating equivalently $\lim_{x\to 0}(\sqrt{x})^{\sqrt{x}}$ which by substitution $x=y^2$ becomes equal to $\lim_{y\to 0}y^y=1$
