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!

Note that $$\lim_{x \to 0}\frac{x}{sinx}=1$$
Therefore $$\frac{1}{2}\lim_{x \to 0}\frac{x}{sinx}=\frac{1}{2}\frac{1}{\lim_{x \to 0}\frac{\sin{x}}{x}}=1/2$$
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.
It's a trivial consequence of algebra of limits (quotient rule). Let $$f(x) =1,g(x)=(\sin x) /x$$ then $$\lim_{x\to 0}\frac{x}{\sin x} =\lim_{x\to 0}\frac{f(x)}{g(x)}=\dfrac{\lim\limits _{x\to 0} f(x)} {\lim\limits _{x\to 0} g(x)} =\dfrac{1}{\lim\limits _{x\to 0} \dfrac{\sin x} {x}}$$ This works because the limit of $$g(x)$$ is non-zero and limit of $$f(x)$$ exists.