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?

Well, we have 
$$
\begin{align}\lim_{x\to 0^+}\sqrt{x}\ln\sqrt{\tan x}&=\lim_{x\to 0^+}\frac{\sqrt{x}}{2}\ln(\tan x)\\
&=\lim_{x\to 0^+}\frac{\ln(\tan x)}{\frac{2}{\sqrt{x}}}\qquad\text{which takes the form } \frac{-\infty}{+\infty}\text{and apply LHR to get}\\
&=\lim_{x\to 0^+}\frac{\frac{\sec^2x}{\tan x}}{-\frac{1}{\sqrt{x^3}}}\\
&=\lim_{x\to 0^+}\frac{\frac{2}{\sin 2x}}{-\frac{1}{x\sqrt{x}}}\\
&=\lim_{x\to 0^+}\bigg[-\sqrt{x}\cdot\frac{2x}{\sin 2x}\bigg]\\
&=-\sqrt{0}\cdot 1=0.
\end{align}$$
Hence, the required limit is 
$$\lim_{x \to 0^+}\sqrt{\tan x}^{\sqrt{x}}=e^{\lim_{x \to 0^+}\ln\sqrt{\tan x}^{\sqrt{x}}}=e^{\lim_{x\to 0^+}\sqrt{x}\ln\sqrt{\tan x}}=e^0=1.$$
