How to prove that $\lim_{x\to 0}\frac{x}{\sin x}=1$, knowing that $\lim_{x\to 0}\frac{\sin x}{x}=1$?

Question

I want to prove $$\lim_{x\to0}\frac{x}{\sin x}=1$$ I know that $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$ Can I just say that I can flip the fraction?

Answer 1

The essential fact that you need here is that the reciprocal function $y\mapsto\dfrac 1 y$ is continuous, and therefore $$ \lim_{x\,\to\,a} \, \frac 1 {g(x)} = \frac 1 {\lim\limits_{x\,\to\,a}\, g(x)}. $$

(In this case, you have $g(x)= \dfrac{\sin x} x.$)

Answer 2

$x→0 ⇒sin x→0 ⇒x≈sinx ⇒ Lim_{x→0}\frac{x}{sinx}=1 $

you can also apply L'Hospital rule:

$Lim_{x→0} \frac{x}{sin x}=\frac{1}{cos 0}=1$