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

The basic limits needed are
$\frac{e^x-1}{x}
\to 1
$,
$\frac{\sin x}{x}
\to 1
$,
and
$\cos x
\to 1
$ as
$x \to 0$.
First,
as $x \to 0$,
$\begin{array}\\
\tan x -\sin x
&=\frac{\sin x}{\cos x}-\sin x\\
&=\sin x(\frac{1}{\cos x}-1)\\
&=\sin x(\frac{1-\cos x}{\cos x})\\
&=\sin x(\frac{2\sin^2(x/2)}{\cos x})\\
&\to x(\frac{2(x/2)^2}{1})\\
&=x^3/2\\
\end{array}
$
Then
$\begin{array}\\
\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}
&=2^{\sin x}\frac{2^{\tan x-\sin x}-1}{x^2\sin x}\\
&\to \frac{2^{\tan x-\sin x}-1}{x^2\sin x}
\qquad\text{since } 2^{\sin x} \to 1\\
&\to \frac{2^{x^3/2}-1}{x^3}\\
&= \frac{e^{\ln 2 x^3/2}-1}{x^3}\\
&\to \frac{\ln 2 x^3/2}{x^3}\\
&= \ln 2/2\\
\end{array}
$
