$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ 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
 A: 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}
$
A: First note that the following can only be used if both resulting limits are convergent

$$\lim\limits_{x\to c}\left[f(x)\pm g(x)\right]=\left[\lim\limits_{x\to c}f(x)\right]\pm \left[\lim\limits_{x\to c}g(x)\right]$$
  $$\lim\limits_{x\to c}\left[f(x)g(x)\right]=\left[\lim\limits_{x\to c}f(x)\right]\left[\lim\limits_{x\to c}g(x)\right]$$

In your first step, you separated the original limit into two divergent limits. Here are the steps in excruciating detail
$$\lim\limits_{x\to 0} \frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}$$
$$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^3}\right]$$
$$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{x^3}\right]$$
$$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x - \sin x}{x^3}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$
$$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{(\sin x)\left(\frac{\tan x}{\sin x} - 1\right)}{x^3}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$
$$=\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\sec x - 1}{x^2}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$
$$=\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\left(\tan x\right)\left(\tan\frac{x}{2}\right)}{x^2}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$
$$=\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x}{x}\right]\left[\frac12\lim\limits_{x\to 0}\frac{\tan\frac{x}{2}}{\frac{x}{2}}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$
Let $s = \frac{x}{2}$, then
$$\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x}{x}\right]\left[\frac12\lim\limits_{s\to 0}\frac{\tan s}{s}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$
Let $t = \tan x - \sin x$, then
$$\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x}{x}\right]\left[\frac12\lim\limits_{s\to 0}\frac{\tan s}{s}\right]\left[\lim\limits_{t\to 0}\frac{2^{t} - 1}{t}\right]$$
$$=\frac12\lim\limits_{t\to 0}\frac{2^{t} - 1}{t}$$
$$=\frac12\log 2$$
A: In a nutshell,
$$\frac{2^{\tan(x)}-2^{\sin(x)}}{x^2\sin(x)}\to\frac{2^{\tan(x)}-2^{\sin(x)}}{\sin^3(x)}\to\frac{2^{(s/\sqrt{1-s^2})-s}-1}{s^3}2^s\to\frac{e^{\ln(2)s^3/2}-1}{s^3}\to\frac{\ln(2)}2.$$



*

*$x$ and $\sin(x)$ are interchangeable (ratio $\to1$),

*$\dfrac1{\sqrt{1-s^2}}-1=\dfrac{1-(1-s^2)}{\sqrt{1-s^2}(1+\sqrt{1-s^2})}\to\dfrac{s^2}2,$

*$\dfrac{e^{at}-1}t\to (e^{at})'|_{t=0}=a$.

A: The fourth line is invalid: If $F(x)$ and $G(x)$ tend to the same limit as $x\to 0,$ we cannot conclude that $(1/x^2)(F(x)-G(x))$ tends to $0.$ For example $F(x)=4 x^2+1$ and $G(x)=x^2+1$. In general, replacing the term $1/x^2$ with any $H(x)$: If $F(x)-G(x)$ and $H(x)$ tend to $0,$ then consider $J(x)=F(x)-G(x).$ We cannot determine whether $J(x)/H(x)$ has a limit, or what the limit is if it exists, just by knowing that $J$ and $H$ tend to $0.$ 
A: @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}
A: One can rewrite:
$$\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\tan x - \sin x}{x^2 \sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{1-\cos x}{x^2 \cos x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\sin^2 x}{x^2}\frac{1}{(1+\cos x)\cos x}$$
$$=2^{\sin x}\cdot\frac{2^{\tan x-\sin x}-1}{\tan x - \sin x}\cdot \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{(1+\cos x)\cos x}\to1\cdot\ln 2\cdot 1^2\cdot \frac{1}{2\cdot1} = \frac{\ln 2}{2}$$
A: 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}
