$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ 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

One can rewrite:
$$\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\tan x - \sin x}{x^2 \sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{1-\cos x}{x^2 \cos x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\sin^2 x}{x^2}\frac{1}{(1+\cos x)\cos x}$$
$$=2^{\sin x}\cdot\frac{2^{\tan x-\sin x}-1}{\tan x - \sin x}\cdot \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{(1+\cos x)\cos x}\to1\cdot\ln 2\cdot 1^2\cdot \frac{1}{2\cdot1} = \frac{\ln 2}{2}$$
