Help find the mistake in this problem of finding limit (using L'Hopital) 
Evaluate $$\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right).$$

Attempt
\begin{align*}
&\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)\\
= &\lim_{x \to 0} \left(\frac{1}{x}-\cot{x}\right)\left(\frac{1}{x}+\cot{x}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right)\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right).
\end{align*}
Both the terms are in $\frac00$ form. So applying L'Hopital on both the limits we have,
$$= \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\sin{x}}{x\cos{x}+\sin{x}}\right).$$
The second term is in $\frac00$ form. So applying L'Hopital on the second limit we have,
\begin{align*}
= &\lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& 1
\end{align*}
The correct answer is $\dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?
 A: As @user21820 encouraged, I am going to explain the errors in this post. 
The problem is the $=$ at the last 2nd line. Before this everything could be accepted, since they are the process to find the limits. But at this step you generally claim that 
$$
\lim f(x) \lim g(x) = \lim f(x)g(x),
$$
where the limits progression are omitted for brevity. By the arithmetic operation of limits, we know that when $\lim f(x), \lim g(x)$ exists, then the equation above holds. In your case, the existence of $\lim f, \lim g$ are not examined at all. Then it is not valid to combine two limits and make it to one limit, since there is a counterexample
$$
1=\lim_{x \to 0} \frac xx , \lim _{x\to 0}x =0, \lim_{x\to 0}\frac 1x \text{ does not exist},
$$
where you obviously cannot write $1 = 0 \times \lim_{x \to 0}(1/x)$.  As other answers showed, not all the two parts 
$$
\lim_{x\to 0} \frac {2\cos x -x \sin x}{x \cos x + \sin x}, \lim_{x\to 0} \frac {x \cos x + \sin x} {2\cos x -x \sin x}
$$
exist, so the certain line makes no sense, and leads you to the wrong result. 
A: \begin{align}\lim_{x\to0}\frac1x+\cot x&=\lim_{x\to0}\frac{\sin x+x\cos x}{x\sin x}\\&=\lim_{x\to0}\frac{2\cos x-x\sin x}{\sin x+x\cos x}\\&=\infty.\end{align}Besides,$$\lim_{x\to0}\frac1x-\cot x=0.$$
The error you made lies in the use of the equality$$\lim_{x\to a}\bigl(f(x)g(x)\bigr)=\left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right)$$in a situation where you cannot apply it (it is the second of your equalities counting from the bottom).
A: 
Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,
$= \lim_{x \to 0} (\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}) \times \lim_{x \to 0}(\frac{x\sin{x}}{x\cos{x}+\sin{x}})$

Note that the limit  $\lim_{x \to 0} (\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}})$ does not exist.
In fact, even before that $\lim_{x \to 0} (\frac{\sin{x}+x\cos{x}}{x\sin{x}})$ does not exist as well.
This is the graph of $\frac{\sin{x}+x\cos{x}}{x\sin{x}}$.

A: Similar work:
$$\begin{align}\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)&=\lim_{x \to 0} \left(\frac{1}{x^2}-\frac{1-\sin^2x}{\sin^2x}\right)=\\
&=\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2x}+1\right)=\\
&=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}\right)+1=\\
&=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}\cdot \frac{\sin^2x}{x^2}\right)+1=\\
&=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)+1.
\end{align}$$
According to the algebraic limit theorem, you can express the limit as a product of two existing limits:
$$\begin{align}\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)+1=&\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^3}\right)}_{-\frac16}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x}\right)}_{=2} +1\stackrel{LR}{=}&\\
\lim_{x \to 0}\left(\frac{\cos x-1}{3x^2}\right)\cdot \lim_{x \to 0}\left(\frac{\cos x+1}{1}\right) +1\stackrel{LR}{=}&\\
\lim_{x \to 0}\left(\frac{-\sin x}{6x}\right)\cdot 2 +1=&\\
-\frac13+1=&\frac23.\end{align}$$
However you can not express:
$$-\frac13=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)=\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}0}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}4}\right)}_{=\infty} \ \ \text{OR}\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}1}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}3}\right)}_{=\infty} \ \ \text{OR}\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}2}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}2}\right)}_{=\infty} \ \ \text{OR}\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}4}\right)}_{=\infty}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}0}\right)}_{=0}.$$
because all are the indeterminate form of $0\cdot \infty$.
A: $$\lim_{x\to0}\left(\dfrac1{x^2}-\dfrac1{\tan^2x}\right)=\lim_{x\to0}\dfrac{\tan x-x}{x^3}\cdot\lim_{x\to0}\dfrac{\tan x+ x}x\cdot\left(\lim_{x\to0}\dfrac x{\tan x}\right)^2$$
Now the limit for the last two are too simple for L'hospital
For $\lim_{x\to0}\dfrac{\tan x-x}{x^3}$ either use L'hospital or  Are all limits solvable without L'Hôpital Rule or Series Expansion
