How to evaluate $\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ I have a stuck on the problem of L'Hospital's Rule, 
$\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ which is in I.F. $\frac{0}{0}$
If we use the rule, we will have 
$\lim\limits_{x\to 0} \frac{\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}}{\sec^2x-\cos x}$.
So, I think that I approach this problem in the wrong way.
Have you guy any idea?

Alternatively, one may use standard Taylor expansions, as $x \to 0$,
$$
\begin{align}
\sin x&=x-\frac{x^3}{6}+o(x^4)
\\\tan x&=x+\frac{x^3}{3}+o(x^4)
\\\arctan x&=x-\frac{x^3}{3}+o(x^4)
\\\arcsin x&=x+\frac{x^3}{6}+o(x^4)
\end{align}
$$ giving, as $x \to 0$,
$$ \frac{\arctan x - \arcsin x}{\tan x - \sin x}= \frac{-\frac{x^3}{2}+o(x^4)}{\frac{x^3}{2}+o(x^4)}=-1+o(x) \to -1.
$$
