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?
Best Answer
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.