Evaluating $\lim_{x\to 0}(\frac{1}{\sin( x)} -\frac{1}{x})$ without L'Hôpital's rule 
Recently, I have been struggling with this one problem:
$$\lim _{x\rightarrow 0}\left(\frac{1}{\sin( x)} -\frac{1}{x}\right)$$
and cannot figure out a way to solve it without the use of L'Hôpital's rule.

The only thing I can think of using is the basic identity
$$\lim_{x\to 0}\left(\frac{\sin( x)}{x}\right) =1$$ but I can't reduce the original problem down to a point where I can apply this identity.
Any help would be greatly appreciated.
Thanks!
 A: \begin{align}
\lim _{x\rightarrow 0}\left(\frac{1}{\sin( x)} -\frac{1}{x}\right)&=\lim _{x\rightarrow 0}\left(\frac{x-\sin(x)}{x^2}\cdot\frac{x}{\sin(x)}\right)\\
&=\lim _{x\rightarrow 0}\left(\frac{x-\sin(x)}{x^2}\right)\\
\end{align}
Now, use the Taylor series of $\sin(x)$.
A: Without L'Hopital
We can add the two fractions together.
$\frac {x - \sin x}{x\sin x}$
And now we need to get a little bit creative.
$|\sin x| \le |x| \le |\tan x|$  (and all have the same sign when $-\frac {\pi}{2} <x< \frac {\pi}{2})$
$|x - \sin x| \le |\tan x - \sin x|$
With this substitution we can find some limits we know how to resolve.
$|\lim_\limits{x\to 0} \frac {x - \sin x}{x\sin x}| \le |\lim_\limits{x\to 0} \frac {\tan x - \sin x}{x\sin x}|$
$\lim_\limits{x\to 0} \frac {\tan x - \sin x}{x\sin x}\\
\lim_\limits{x\to 0} \frac {\sec x - 1}{x}\\
\lim_\limits{x\to 0} \left(\frac {1 - \cos x}{x}\right) \sec x = 0$
And by the squeeze theorem.
$0\le |\lim_\limits{x\to 0} \frac {1}{\sin x} - \frac {1}{x}| \le \sec x \lim_\limits{x\to 0} \frac {1 - \cos x}{x}$
Gives us $0.$
A: You could use the first few terms of the Taylor series for $\sin(x)$ around $x = 0$ (i.e., the Maclaurin series) in the numerator to get
$$\begin{equation}\begin{aligned}
\lim_{x \rightarrow 0}\left(\frac{1}{\sin( x)} - \frac{1}{x}\right) & = \lim_{x \rightarrow 0}\left(\frac{x - \sin(x)}{x\sin(x)}\right) \\
& = \lim_{x \rightarrow 0}\left(\frac{x - (x - \frac{x^3}{3} + O(x^5))}{x\sin(x)}\right) \\
& = \lim_{x \rightarrow 0}\left(\frac{\frac{x^2}{3} + O(x^4))}{\sin(x)}\right) \\
& = \lim_{x \rightarrow 0}\left(\frac{x}{\sin(x)}\right)\left(\frac{x}{3} + O(x^3))\right) \\
& = 1(0) \\
& = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
