$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong? 
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions).
The actual answer for the given question is $\frac{1}{2}\log(2)$.
My course book has shown that don't use this step but has not given the reason.
AND Please TELL why i am WRONG

Let $u=\tan \frac x2$ and kill: 
\begin{align}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}&=\frac{2^{\frac{2u}{1-u^2}} - 2^{\frac{2u}{1+u^2}}}{\frac{8u}{1+u^2}\arctan^2u }\\
&=(1+u^2)2^{\frac{2u}{1+u^2}}\frac{2^{\frac{4u^3}{1-u^4}} - 1}{8u\arctan^2u }\\
\end{align}
Now since $$\frac{4u^3}{1-u^4}=4\sin^3\frac x2 \times \frac{\cos \frac x2}{\cos x}$$ we have 
\begin{align}\lim _{x \to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}
&=\lim _{u \to 0}(1+u^2)2^{\frac{2u}{1+u^2}}\times \lim _{u \to 0}\frac{2^{\frac{4u^3}{1-u^4}} - 1}{8u\arctan^2u }\\
&=\lim _{u \to 0}\frac{2^{\frac{4u^3}{1-u^4}} - 1}{8u\arctan^2u }\\
&=\lim _{x \to 0}\frac{2^{4x^3} - 1}{8x^3 }\\
&=\frac18\times \ln 2^4
\end{align}
