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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
$$frac{1}{x}-frac{1}{sin x}=frac{sin x-x}{xsin x}.$$
Let

$$f(x)=frac{x-sin x}{xsin x}$$
(the change of sign is for convenience). We will show that $limlimits_{xto 0},f(x)=0.$

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 $x-sin x$:
$$tan x-sin x=sin xleft(frac{1-cos x}{cos x}right)=sin xleft(frac{1-cos^2 x}{cos x(1+cos x)}right)=frac{sin^3 x}{cos x(1+cos x)}.$$
We conclude that
$$0 <frac{x-sin x}{xsin x}<frac{sin^2 x}{xcos x(1+cos x)}.$$
Since $sin x<x$, we find that
$$0 <frac{x-sin x}{xsin x}<frac{sin x}{cos x(1+cos x)},$$
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.

Simplify to have $$frac{sin x-x }{xsin x}$$ 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 $xrightarrow 0$, we have;

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

which is the required answer.

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

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

If you believe (or know how to show) that the function $displaystyle{f(x)=frac{x}{sin(x)}}$, $xneq 0$, $f(0)=1$ 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 $-f'(0)=0$.

METHOD I

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

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

Q.E.D.

METHOD II

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

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

Q.E.D.

Using $$sin x<x<tan xqquad(0<x<{piover2})$$ we have
$${sin(x/2)over x/2} {sin(x/2)overcos x}={1-cos xover x>cos x}>{1overtan x>cos x}-{1over x}={1oversin x}-{1over x}>0qquad(0<x<{piover2}) .$$
Letting $xto0+$ the left hand side converges to $0$ because of $lim_{tto0}{sin tover t}=1$.

I did it that way:$$lim_{xto0} left(frac{1}{x} - frac{1}{sin x}right) =lim_{xto0} left(frac{1}{x} - frac{1}{frac{sin x}{x}*{x}}right) $$
because $lim_{xto0} frac{sin x}{x} = 1$ then
$$ lim_{xto0} left(frac{1}{x} - frac{1}{frac{sin x}{x}*{x}}right) \= lim_{xto0} left(frac{frac{sin x}{x}* x - x}{frac{sin x}{x}*x}right)\= lim_{xto0} left(frac{x(frac{sin x}{x}* 1 - 1)}{frac{sin x}{x}*x}right) \=lim_{xto0} left(frac{frac{sin x}{x} - 1}{frac{sin x}{x}}right) =[frac{0}{1}] = 0 $$