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Find the limit:

$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$

I am not able to find it because I don't know how to prove or disprove $0$ is the answer.

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@JonasMeyer - Thanks –  Victor Dec 29 '11 at 3:56
Do you have calculus at your disposal? –  Jonas Meyer Dec 29 '11 at 3:58
This is an odd function, so there cannot be a finite non-zero limit at $0$. –  Henry Dec 29 '11 at 12:01

8 Answers 8

up vote 12 down vote accepted

Hint: Try using $$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\left(\frac{\sin x - x}{x\sin x}\right)$$ and apply L'Hopital's rule.

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So, is the answer is just (cos x -1)/(sin(x) + xcos(x)) or can be futher simplifly? –  Victor Dec 29 '11 at 4:07
A second application gives $\lim_{x\to 0} -\sin x/(2\cos x - x\sin x)$, which should be solvable. –  Alex Becker Dec 29 '11 at 4:14

For fun, and because of the pre-calculus tag, we give a proof without calculus. It turns out that there is a geometric argument that $|x-\sin x|$ is less than a constant times $|x^3|$ for $x$ near $0$.

I will need some help from you, to draw the missing picture. We have $$\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x}.$$ Let
$$f(x)=\frac{x-\sin x}{x\sin x}$$ (the change of sign is for convenience). We will show that $\lim\limits_{x\to 0}\,f(x)=0.$

We are interested in the behaviour of $f(x)$ when $x$ is close (but not equal) to $0$. Note that $f(-x)=-f(x)$. So we will be finished if we can show that $f(x)$ approaches $0$ as $x$ approaches $0$ through positive values.

Let $x$ be small positive. Draw $\triangle OPQ$ as follows. The base of the triangle is $OP$, and has length $1$. The triangle is right-angled at $P$. Finally, $Q$ is such that $\angle QOP =x$.

Draw the circular sector with centre $O$, radius $1$, and going from $P$ to a point on $OQ$. So the sector has angle $x$.

Note that the circular sector is contained in $\triangle OPQ$. The circular sector has area $(1/2)x$, and $\triangle OPQ$ has area $(1/2)\tan x$. Thus the geometry gives us the inequality $$\frac{x}{2}<\frac{\tan x}{2}.$$ Since $x>\sin x$, we get the estimates $$0<x-\sin x< \tan x-\sin x.$$ The right-hand side only involves trigonometric functions, so is easier to deal with than $x-\sin x$: $$\tan x-\sin x=\sin x\left(\frac{1-\cos x}{\cos x}\right)=\sin x\left(\frac{1-\cos^2 x}{\cos x(1+\cos x)}\right)=\frac{\sin^3 x}{\cos x(1+\cos x)}.$$ We conclude that $$0 <\frac{x-\sin x}{x\sin x}<\frac{\sin^2 x}{x\cos x(1+\cos x)}.$$ Since $\sin x<x$, we find that $$0 <\frac{x-\sin x}{x\sin x}<\frac{\sin x}{\cos x(1+\cos x)},$$ and it is clear that $\dfrac{\sin x}{\cos x(1+\cos x)}$ approaches $0$ as $x$ approaches $0$ through positive values.

Comment: In this problem, there is no virtue in avoiding the calculus. The Taylor expansion is the natural approach.

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Simplify to have $$\frac{\sin x-x }{x\sin x}$$ and consider Maclaurin's series for $$\sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}-...$$

So you have $$\frac{(x-\frac {x^3}{3!}+\frac {x^5}{5!}-...)-x}{x(x-\frac {x^3}{3!}+\frac {x^5}{5!}+...)}=\frac{(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}.$$

Finding the limit as $x\rightarrow 0$, we have;

$$\frac{\lim_{x\rightarrow 0}(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{\lim_{x\rightarrow 0}(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}=\frac{0}{1}=0.$$

which is the required answer.

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Since everybody was 'clever', I thought I'd add a method that doesn't really require much thinking if you're used to asymptotics.

The power series for $\sin x$

$$\sin x = x + O(x^3)$$

We can compute the inverse of this power series without trouble. In great detail:

$$\begin{align}\frac{1}{\sin x} &= \frac{1}{x + O(x^3)} \\ &= \frac{1}{x} \left( \frac{1}{1 - O(x^2))} \right) \\ &= \frac{1}{x} \left(1 + O(x^2) \right) \\ &= \frac{1}{x} + O(x) \end{align}$$

going from the second line to the third line is just the geometric series formula. Anyways, now we can finish up:

$$\frac{1}{x} - \frac{1}{\sin x} = O(x)$$

$$ \lim_{x \to 0} \frac{1}{x} - \frac{1}{\sin x} = 0$$

If we wanted, we could get more precision: it's not hard to use the same method to show

$$ \frac{1}{\sin x} = \frac{1}{x} + \frac{x}{6} + O(x^3) $$

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+1 for the cleverness-free approach. –  Did Aug 16 '12 at 11:03

If you believe (or know how to show) that the function $\displaystyle{f(x)=\frac{x}{\sin(x)}}$, $x\neq 0$, $f(0)=1$ is differentiable at $0$, then because $f$ is even, it follows that $f'(0)=0$. Note that $\frac{1}{x}-\frac{1}{\sin(x)}=-\frac{f(x)-f(0)}{x}$, so the limit in question is $-f'(0)=0$.

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Can you prove it? I think if I can prove the derivative is continuous, then it is done? But I cannot prove that either. –  Sawarnik Feb 25 '14 at 10:11


Firstly, notice that the expression under the limit is an odd function and consider that $\sin(x)<x$. Then we have that: $$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\frac{\sin x - x}{x\sin x}\le\lim_{x \rightarrow 0}\frac{\sin x - x}{x^2}\le\lim_{x \rightarrow 0}\frac{\tan x - x}{x^2}=0$$

As regards the last limit you wanna see my proof here.



$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\frac{\sin x - x}{x\sin x}\le\lim_{x \rightarrow 0}\frac{\sin x - x}{x^2}=\lim_{x \rightarrow 0}x\cdot\frac{\sin x - x}{x^3}=0\cdot-\frac{1}{6}=0$$

Let's solve now the auxiliary limit I used (elementarily): $$L=\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}=\lim_{x \rightarrow 0}\frac{\sin 2x - 2x}{8x^3}=\lim_{x \rightarrow 0}\frac{\sin x \cos x - x}{4x^3}=\lim_{x \rightarrow 0}\frac{\sin x \cos x -x\cos x + x\cos x- x}{4x^3}=\lim_{x \rightarrow 0}\frac{\cos x(\sin x \ -x) }{4x^3}-\lim_{x \rightarrow 0}\frac{(1 - \cos x) }{4x^2}=$$ $$\lim_{x \rightarrow 0} \cos x \cdot\frac{L}{4} -\frac{1}{8}=\frac{L}{4}-\frac{1}{8}$$ $$L=\frac{L}{4}-\frac{1}{8}$$ $$L=-\frac{1}{6}.$$


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But $\frac{\sin x - x}{x\sin x} \le \frac{\sin x - x}{x^2}$ does not follow from $\sin(x) < x$. –  Teddy Nov 11 '12 at 16:39

Using $$\sin x<x<\tan x\qquad(0<x<{\pi\over2})$$ we have $${\sin(x/2)\over x/2}\ {\sin(x/2)\over\cos x}={1-\cos x\over x\>\cos x}>{1\over\tan x\>\cos x}-{1\over x}={1\over\sin x}-{1\over x}>0\qquad(0<x<{\pi\over2})\ .$$ Letting $x\to0+$ the left hand side converges to $0$ because of $\lim_{t\to0}{\sin t\over t}=1$.

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I did it that way:$$\lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\sin x}\right) =\lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\frac{\sin x}{x}*{x}}\right) $$ because $\lim_{x\to0} \frac{\sin x}{x} = 1$ then $$ \lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\frac{\sin x}{x}*{x}}\right) \\= \lim_{x\to0} \left(\frac{\frac{\sin x}{x}* x - x}{\frac{\sin x}{x}*x}\right)\\= \lim_{x\to0} \left(\frac{x(\frac{\sin x}{x}* 1 - 1)}{\frac{\sin x}{x}*x}\right) \\=\lim_{x\to0} \left(\frac{\frac{\sin x}{x} - 1}{\frac{\sin x}{x}}\right) =[\frac{0}{1}] = 0 $$

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BTW you can use \cdot instead of *, for example $\frac{\sin x}x\cdot x$ is typeset as $\frac{\sin x}x\cdot x$. –  Martin Sleziak 20 hours ago

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