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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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    See [What is the result of $\lim_{$x$ \rightarrow 0}\left(\frac1$x$ - \frac1{\sin x}\right)$?](htt$p$://math.stackexchange.com/questions/94864/what-is-the-result-of-lim-x-rightarrow-$0$-left-frac1x-frac1-sin-x-righ)2012-08-15

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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.