<s>[INST] How to find $\lim_{x \to 0} ( \frac{1}{\sin(x)}- \frac{1}{\arcsin(x)})$ I want to do the problem without using L'Hopitals rule, I have
$$\frac{1}{\sin(x)}- \frac{1}{\arcsin(x)} = \frac{x}{\sin(x)}\frac{x}{\arcsin(x)}\frac{\sin(x)-\arcsin(x)}{x^2}$$
and I'm not quite sure about how to deal with the $\dfrac{\sin(x)-\arcsin(x)}{x^2}$, apparently its limit is $0$? In which case the whole limit would be $0$. But how would I show this without using l'Hopitals rule. Thanks for any help.
 [/INST] Using Taylor's theorem:
$$\color{blue}{\sin x=x-x^3/6+x^5/120+O(x^7)}$$
$$\color{red}{\arcsin x=x+x^3/6+3x^5/40+O(x^7)}$$
So:
$$\begin{align}\left( \frac{1}{\color{blue}{\sin(x)}}- \frac{1}{\color{red}{\arcsin(x)}}\right)&=\frac{\color{red}{\arcsin x}-\color{blue}{\sin x} }{\color{blue}{\sin x}\arcsin x}\\&=\frac{\color{blue}{\color{red}{(x+x^3/6+O(x^5))}-(x-x^3/6+O(x^5))}}{\color{blue}{(x-x^3/6+O(x^5))}\color{red}{(x+x^3/6+O(x^5))}}\\&=\frac{+x^3/3+O(x^5)}{x^2+O(x^6)}\end{align}$$
So:
$$\lim \limits_{x \to 0}\left( \frac{1}{\sin(x)}- \frac{1}{\arcsin(x)}\right)=0$$
</s>