<s>[INST] Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$ 
Evaluate  $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$

It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
 [/INST] Notice that the limit is
$$
L=\lim_{x \to 0}\frac{1}{\sin x} \left(\frac{1}{\arctan x} - \frac{\cos x}{\arcsin x} \right)
$$
from the series expansion for small values of $x$ one has $\cos x =1- \frac{x^2}{2} + O(x^4)$ and $(\mbox{arc})\sin x = x + O(x^3) = \arctan x$ hence
$$L = \lim_{x \to 0}\frac{1}{(x + O(x^3))} \left(\frac{1}{x + O(x^3) } - \frac{1-x^2/2 + O(x^4)}{x + O(x^3)} \right)=\lim_{x \to 0} \frac{1}{x}\left(\frac{x}{2} + O(x^3)  \right) = \frac{1}{2}$$
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