# Limit of $\frac{1}{x} - \frac{1}{\sin{(x)}}$ [duplicate]

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.

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jon jones: 0% accept rate... Why? –  Did Aug 15 '12 at 16:22
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## marked as duplicate by Martin Sleziak, Did, Ｊ. Ｍ., DonAntonio, Ross MillikanAug 15 '12 at 16:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

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.

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