<s>[INST] 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?
 [/INST] The first step is OK, but you should try and simplify things before going on.
The numerator is
$$
\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}=
\frac{\sqrt{1-x^2}-1-x^2}{(1+x^2)\sqrt{1-x^2}}=
\frac{-x^2(3+x^2)}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)}
$$
The denominator is
$$
\frac{1}{\cos^2x}-\cos x=\frac{1-\cos^3x}{\cos^2x}=
\frac{(1-\cos x)(1+\cos x+\cos^2x)}{\cos^2x}
$$
so you can rewrite your limit as
$$
\lim_{x\to0}
\frac{x^2}{1-\cos x}
\frac{-(3+x^2)\cos^2x}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)(1+\cos x+\cos^2x)}
$$
which is easy to compute (the second fraction is not “indeterminate”).
Of course, Taylor expansion is easier.
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