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

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.

    
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
    
Your solution is already very nice. –  vadim123 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

4 Answers 4

up vote 2 down vote accepted

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

so that

$$\frac{1}{x}-\frac{1}{\sin{x}} \sim-\frac{x}{6}$$

The limit is zero.

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