What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$? 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.
 A: 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
$$ \frac{1}{\sin x} = \frac{1}{x} + \frac{x}{6} + O(x^3) $$
A: 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 $-f'(0)=0$.
A: METHOD I
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.
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.
A: 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 $\lim_{t\to0}{\sin t\over t}=1$.
A: 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.   
A: 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.
A: 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.
A: 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
$$ \lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\frac{\sin x}{x}*{x}}\right) \\= \lim_{x\to0} \left(\frac{\frac{\sin x}{x}* x - x}{\frac{\sin x}{x}*x}\right)\\= \lim_{x\to0} \left(\frac{x(\frac{\sin x}{x}* 1 - 1)}{\frac{\sin x}{x}*x}\right) \\=\lim_{x\to0} \left(\frac{\frac{\sin x}{x} - 1}{\frac{\sin x}{x}}\right) =[\frac{0}{1}] = 0 $$
