# Elegant or elementary evaluation of $\lim\limits_{x\to 0} \left( \frac{1}{x}-\frac{1}{\sin(x)} \right)$ [duplicate]

I give math tutoring and was wondering about the following limit. I found the answer but I was wondering if someone has a nicer explanation than the one I am giving where I use L'Hôpital's rule twice. The limit I am evaluating is: $$\lim\limits_{x\to 0} \left( \frac{1}{x}-\frac{1}{\sin(x)} \right) = \lim\limits_{x\to 0} \frac{\sin(x)-x}{x\sin(x)}$$ Both numerator and denomitor go to zero here so we may use l'Hôpital's rule here. The limit is thus equal to: $$\lim\limits_{x\to 0} \frac{\cos(x)-1}{\sin(x)+x\cos(x)}$$ Again numerator and denominator go to zero and thus by l'Hôpital the limit is equal to: $$\lim\limits_{x\to 0} \frac{-\sin(x)}{\cos(x) + \cos(x) + x\sin(x)}$$ Now we are looking at the limit of the quotient of two everywhere continuous functions where the denominator is not zero. Thus the function itself is continuous and so the limits is: $$\lim\limits_{x\to 0} \frac{-\sin(x)}{2 \cos(x) + x\sin(x)} = \frac{0}{2}=0$$

Does anyone know a more nicer, more elementary way of solving this? Thanks!

EDIT: Also, if anyone knows a fast, yet less elementary way to solve I would enjoy seeing it so feel free to post :)

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You can use Taylor expansions also. Knowing $\sin(x)=x-\frac{x^3}{6}+o(x^4)$ suffices to compute the limit. –  Ewan Delanoy Oct 3 '13 at 13:49
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. –  Lord_Farin Oct 3 '13 at 13:59
Thanks, I will take this into account in the future! –  Slugger Oct 3 '13 at 14:04

## marked as duplicate by njguliyev, Julian Kuelshammer, Old John, Macavity, SeiriosOct 3 '13 at 14:23

Let us first look at what happens when $x > 0$. For $x \in ( 0, \frac{\pi}{2} )$, we have

$$0 < \sin x < x < \tan x \implies 0 < \frac{1}{\sin x} - \frac{1}{x} < \frac{1}{\sin x} - \frac{1}{\tan x} = \frac{1 - \cos x}{\sin x} = \tan\frac{x}{2}$$

Since $\tan \frac{x}{2} \to 0$ as $x \to 0^{+}$, we have $$\lim_{x\to 0^{+}} \left( \frac{1}{x} - \frac{1}{\sin x} \right) = 0$$

Since both $x$ and $\sin x$ are odd functions in $x$, this immediately implies

$$\lim_{x\to 0^{-}} \left( \frac{1}{x} - \frac{1}{\sin x} \right) = 0$$

and hence the $x \to 0$ limit exists and equal to $0$.

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If you're allowed/want to use power series

$$\frac1x-\frac1{\sin x}=\frac1x-\frac1{x\left(1-\frac{x^2}6+\ldots\right)}=$$

$$=\frac1x-\frac1x\left(1+\frac{x^2}{6}+\frac{x^4}{6^2} +\ldots\right)=-\frac{x}6-\ldots\xrightarrow[x\to 0]{}0$$

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$$\left( \frac{1}{x}-\frac{1}{\sin(x)} \right) = \frac{1}{x} -\frac{1}{x-x^3/6+O(x^3)} = \frac{1}{x}-\frac{1}{x}.\frac{1}{1-x^2/6+O(x^2)}$$ $$= \frac{1}{x} - \frac{1}{x}.(1+x^2/6+O(x^2))$$ $$= -\frac{x}{6}+O(x)$$

thus $$\lim \limits_{x \to 0} \left( \frac{1}{x}-\frac{1}{\sin(x)} \right) = 0$$

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$$\frac{1}{x}-\frac{1}{\sin{x}} \sim \frac{1}{x}-\frac{1}{x-x^3/6+\cdots} = \frac{1}{x} - \frac{1}{x} \frac{1}{1-x^2/6+\cdots}=\frac{1}{x}-\frac{1}{x}\left ( 1+\frac{x^2}{6}+\cdots\right)$$
$$\frac{1}{x}-\frac{1}{\sin{x}} \sim-\frac{x}{6}$$