How to evaluate the limit $\lim_\limits{x\to 0+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right )$?

Question

$$\lim_{x\to 0^+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right ) =\ ?$$

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?

Answer 1

Write it as

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

Answer 2

If you use L'hopital's rule two more times, you will reach the answer. Though using the finite expansion sin(x) ~ x - x^3/3! makes it a lot more easier. The limit will be zero. (Edited).

Answer 3

With MacLaurin formulas it's straightforward: $$\lim_{x\to 0^+}\frac1{\sqrt{x}}\left(\frac1{\sin x} - \frac1{x}\right) = \lim_{x\to 0^+}\frac{x - \sin x}{x\sqrt x\sin x} = \lim_{x \to 0^+}\frac{\frac16x^3 + o(x^3)}{x^{5/2} + o(x^{5/2})} = \lim_{x \to 0^+}\frac{\sqrt x}6 = 0$$

We used the fact that $$\sin x = x - \frac{x^3}{3!} + o(x^3)$$