evaluate $\lim_{x \to 0+} \frac{x-\sin x}{(x \sin x)^{3/2}}$

Question

Evaluate $\lim_{x \to 0+} \frac{x-\sin x}{(x \sin x)^{3/2}}$

This is an exercise after introducing L'Hopital's rule. I directly apply L'Hopital's rule three times and it becomes more and more complex. So I try to substitute $t=\sqrt x$ but there's a square root remained in the denominator $t^3 (sin(t^2))^{3/2}$,apply L'Hopital's rule three times or more I still can't solve it. So I think maybe I have to write $t=\sqrt{x \sin x}$ ,but I can't find a way to convert $x- \sin x$ to a function about $t$ .

Thanks in advance.

Answer 1

$$\begin{align*} \lim_{x\to0^+}\frac{x-\sin x}{(x\sin x)^{3/2}}&=\lim_{x\to0^+}\left(\frac{x}{\sin x}\right)^{3/2}\frac{x-\sin x}{x^3}\\ &=\lim_{x\to0^+}\frac{x-\sin x}{x^3} \end{align*}$$

And apply L'Hopital's rule three times.

Answer 2

By Taylor series we have $$\sin x\sim_0 x-\frac{x^3}{6}$$ hence $$\lim_{x \to 0+} \frac{x-\sin x}{(x \sin x)^{3/2}}=\lim_{x \to 0+}\frac{x^3/6}{(x^2)^{3/2}}=\frac16$$