Evaluate $\lim_{x \to 0} \frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$ Evaluate $$\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$$
First I tried using L'Hopital's rule..but it's very lengthy
Next I have written the limits as
$$L=\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}}{\lim_{x \to 0}\frac{\tan x-\sin x}{x^3}}=\frac{L_1}{L_2}$$
Now by L'Hopital's Rule we get $L_2=0.5$
$$L_1=\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}$$
Now $L_1$ can also be evaluated using three applications of L'Hopital's Rule, but is there any other approach?
 A: \begin{align}
\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}&= \lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)\cos(\sin x)}{\tan x - \sin x}\\
&=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)+\tan(\sin x)-\tan(\sin x)\cos(\sin x)}{\tan x -\sin x}\\
&=\lim_{x\to 0}\left(\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}+\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}\right)
\end{align}
Then, by mean value theorem, there is $c\in (-\frac{\pi}{2},\frac{\pi}{2})$ such that
$$
\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\sec^2 c
$$
and between $\sin x$ and $\tan x$. Then $\lim_{x\to 0} \frac{\tan(\tan x)-\tan(\sin x)}{\tan x - \sin x}=1$. Next, we will show that $\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}=1$.
\begin{align}
\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}&=\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))\cos x}{\sin x(1 - \cos x)}\\
&=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\cdot \frac{1-\cos(\sin x)}{\sin^2 x}\cdot \frac{\sin^2 x}{1-\cos x}\cdot \cos x\\
&=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\lim_{x\to 0}\frac{1-\cos(\sin x)}{\sin^2 x}\lim_{x\to 0}(1+\cos x)\lim_{x\to 0}\cos x\\
&=1\cdot \frac{1}{2}\cdot 2\cdot 1\\
&=1.
\end{align}
Therefore,
\begin{align}
\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}&=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)}{\tan x - \sin x}+\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}\\
&=1+1\\
&=2
\end{align}
