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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.

  • 10
    This is an odd function, so there cannot be a finite non-zero limit at $0$.2011-12-29

8 Answers 8

-1

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 $

  • 0
    BTW you can use `\cdot` instead of `*`, for example $\frac{\sin x}x\cdot x$ is typeset as `$\frac{\sin x}x\cdot x$`.2015-08-02
17

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 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, 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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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.

  • 3
    A second application gives $\lim_{x\to 0} -\sin x/(2\cos x - x\sin x)$, which should be solvable.2011-12-29
10

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) $

  • 1
    +1 for the [cleverness-free](http://math.stackexchange.com/a/66416/6179) approach.2012-08-16
4

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.2014-02-25
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METHOD I

Firstly, notice that the expression under the limit is an odd function and consider that $\sin(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.

Q.E.D.

METHOD II

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

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

Q.E.D.

  • 0
    But $\frac{\sin x - x}{x\sin x} \le \frac{\sin x - x}{x^2}$ does not follow from \sin(x) < x.2012-11-11
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Using $\sin x 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 Letting $x\to0+$ the left hand side converges to $0$ because of $\lim_{t\to0}{\sin t\over t}=1$.