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

Question

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 1

Note that $$ \lim_{x \to 0}\frac{x}{sinx}=1$$

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

Answer 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.

Answer 3

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.