[Math] How to evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L’Hopital or series

Question

How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?

I've tried expanding the variable such as $x = 2y$ or $x = 3y$, but seemed to still get stuck.

Answer 1

Once we've proven $\lim_{x\to0}\frac{\sin x}{x}=1$ and $\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac16$ (see e.g. here and here for solutions using neither L'Hôpital's rule nor series), your limit is $\lim_{x\to0}\frac{x-\sin x}{x^3}\frac{x^2}{\sin x}=0$.