calculus - Why does $frac{1}{2}lim_{x to 0}frac{x}{sin x}$ equal to $frac{1}{2}frac{1}{lim_{x to 0}frac{sin{x}}{x}}$?

I've come across the following transformation:

$$\frac{1}{2}\lim_{x \to 0}\frac{x}{\sin x}=\frac{1}{2}\frac{1}{\lim_{x \to 0}\frac{\sin{x}}{x}}$$

But I can't quite understand why and how it works. I would be grateful if someone explained why it's correct.

Thanks!

Answer

The function $x\mapsto \frac1x$ is continuous. Therefore, for any function $f(x)$ and any value $a\in[-\infty,\infty]$, we have $$\lim_{x\to a}\frac1{f(x)}=\frac1{\lim_{x\to a}f(x)}$$as long as any of the expressions exist.