<s>[INST] How to evaluate the limit $\lim_\limits{x\to 0+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right )$? $$\lim_{x\to 0^+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right ) =\ ?$$
I rearranged it as
$$\lim_{x\to 0^+ } \frac{x-\sin x}{x\sqrt{x}\sin x} = \lim_{x\to0^+ } \frac{x-\sin x}{x^{\frac{3}{2}}\sin x}$$
Which gives an indetermination of the form $0/0$. Then, I tried L'Hospital:
$$\lim_{x\to 0^+ } \frac{x-\sin x}{x^{\frac{3}{2}}\sin x} = \lim_{x\to 0^+ } \frac{1-\cos x}{\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}} \cos x}$$
Should I continue to apply L'Hospital or is there a simpler way to solve it?
 [/INST] Write it as

$$ \frac{x-\sin x}{x^{3/2}\sin x} \sim_{x\sim 0} \frac{x^3/3!}{x^{3/2}\,x} = \frac{1}{3!}x^{1/2}\longrightarrow_{x\to 0}0. $$

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