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?

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
$$\sin x = x - \frac{x^3}{3!} + o(x^3)$$
