Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$ Find lim: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$$.
You can use L'Hospitale, or Maclaurin, etc
 A: Hint: Use $\tan x = x+\dfrac{x^3}{3}+o(x^3)$ and $\sin x = x-\dfrac{x^3}{6}+o(x^3)$ to obtain $\tan(\tan x) = x+\dfrac{2x^3}{3}+o(x^3)$ and $\sin (\sin x) = x-\dfrac{x^3}{3}+o(x^3)$.

 Hence $$\dfrac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x} = \dfrac{x^3+o(x^3)}{\frac{x^3}{2}+o(x^3)} \to 2.$$

A: Assume that $h(x)\to0$ when $x\to0$ and that $f$ and $g$ are such that $$
f(x)=h(x)+ah(x)^{1+n}+o(h(x)^{1+n}),
$$
and
$$
g(x)=h(x)+bh(x)^{1+n}+o(h(x)^{1+n}),
$$ 
with $a\ne b$, for some $n\gt0$. Then, for every $i\geqslant1$, 
$$
f^{\circ i}(x)=h(x)+iah(x)^{1+n}+o(h(x)^{1+n}),
$$
and
$$
g^{\circ i}(x)=h(x)+ibh(x)^{1+n}+o(h(x)^{1+n}),
$$ hence
$$
f^{\circ i}(x)-g^{\circ i}(x)\sim i(a-b)h(x)^{1+n},
$$
and, for every positive $i$ and $j$,
$$
\lim_{x\to0}\frac{f^{\circ i}(x)-g^{\circ i}(x)}{f^{\circ j}(x)-g^{\circ j}(x)}=\frac{i}j.
$$
Application: for $h(x)=x$, $f(x)=\tan x$, $g(x)=\sin x$ hence $n=2$, $a=\frac13\ne-\frac16=b$, choose $i=2$ and $j=1$.
A: One way without MacLaurin or L'Hospital: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}=\lim_{x\to0} \frac{\tan(\tan x) - \tan(\sin x)+\tan(\sin x)-\sin(\sin x)}{\tan x -\sin x} = \lim_{x\to0} \frac{[\tan(\tan x -\sin x)][1-\tan(\tan x)\tan(\sin x)]}{\tan x -\sin x}+\lim_{x\to0} \frac{\sin(\sin x)[1 - \cos(\sin x)]}{\cos(\sin x)(\tan x -\sin x)}=$$ $$=1+\lim_{x\to0} \frac{\sin(\sin x)}{\sin x}\lim_{x\to0} \frac{1 - \cos(\sin x)}{\sin^2 x}\lim_{x\to0}\frac{x^2}{1-\cos x}\lim_{x\to0}\frac{\sin^2x}{x^2} = 1+1.\frac{1}{2}.2.1 =2  $$.
We used $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b}$$ $$\lim_{x\to0}\frac{\sin x}{x}= \lim_{x\to0}\frac{\tan x}{x}=\lim_{x\to0}\cos x=1, \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2} $$
