This is an exercise from the book "Differential and Integral Calculus" by N. Piskunov, page 161, exercise 68.
$$\lim\limits_{x\rightarrow 0} (\frac{1}{x^2} – \frac{\cot(x)}{x}) = \frac{1}{3}$$.
My first idea was to rewrite this limit as:
$$\lim\limits_{x\rightarrow 0} \frac{1-x\cot(x)}{x^2}$$.
And I defined the function $f(x) = 1 – x\cot(x)$ to try and get a Maclaurin expansion for it and then evaluate the limit. The problem lies in the fact that the first derivative of $f(x)$, $f^{\prime}(x) = \frac{x^2\csc^2(x)+x
\cot(x)-2}{x^3}$ which is undefined at $x = 0$.
Best Answer
It's so strange about the fact that $f'$ is undefined at $0$? After all, $f$ itself is undefined at $0$.
On the other hand$$\frac1{x^2}-\frac{\cot(x)}x=\frac{\sin(x)-x\cos(x)}{x^2\sin(x)}$$and you have$$\sin(x)-x\cos(x)=\frac{x^3}3+O(x^4)\quad\text{and}\quad x^2\sin(x)=x^3+O(x^4).$$So, your limit is equal to$$\lim_{x\to0}\frac{x^3/3}{x^3}=\frac13.$$