<s>[INST] 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?
 [/INST] 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 $0\cdot \infty$.
</s>