<s>[INST] $\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ 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
 [/INST] @vim suggested Taylor expansions:
\begin{align}\lim _{x \to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}
&=\lim _{x \to 0}\frac{2^{x+\frac{x^3}{3}+O(x^5)}-2^{x-\frac{x^3}{6}+O(x^5)}}{x^2(x+\frac{x^3}{3}+O(x^5))}\\
&=\lim _{x \to 0}\frac{\left(2^{x+O(x^5)}\right)\left(2^{\frac{x^3}{3}}-2^{\frac{x^3}{6}}\right)}{x^3\left(1+O(x^5)\right)}\\
&=\lim _{x \to 0}\frac{2^{x-\frac{x^3}{6}+O(x^5)}}{1+O(x^5)}\cdot\lim_{x\to0}\frac{2^{\frac{x^3}{2}} - 1}{x^3 }\\
&=1\cdot\lim_{x\to0}\frac12\cdot\frac{2^{\frac{x^3}{2}} - 1}{\frac{x^3}{2} }\\
&=\frac12\cdot\lim_{u\to0}\cdot\frac{2^{u} - 1}{u }\\
&=\frac12\cdot \ln 2
\end{align}
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