Limit of $\frac{1}{x} - \frac{1}{\sin{(x)}}$

Question

Prove, without using l'Hôpital's Rule, that $\lim\limits_{x \to 0}{\dfrac{1}{x} - \dfrac{1}{\sin{(x)}}} = 0$.

I proved that there exists a $s >0$ such that $\forall x \in (-s,s)$ $\Rightarrow$ $\dfrac{1}{x} - \dfrac{1}{\sin{(x)}} > 0$ if $x<0$ and $\dfrac{1}{x} - \dfrac{1}{\sin{(x)}} < 0$ if $x>0$. Therefore, this limit exist and it is equal zero, or doesn’t exist. But it is only thing I could do.

Answer 1

Write $ \lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$ as $ \lim_{x \rightarrow 0}\left(\frac{\sin x - x}{x\sin x}\right)$ and then expand $\sin x$. Your observation is correct.