# Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$

I solved it by rewriting the limit as indeterminate form $$0/0$$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!

\begin{align} \lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})&=\lim_{x\to0^+}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2 x}\\ \vdots\\ &=\lim_{x\to0^+}\frac{-16\cos 2x+\ldots}{24\cos 2x+\ldots}\\ &=-\frac{2}{3} \end{align}

• Desired limit is same as that of $(x^2-\tan^2x)/x^4$. Split this using $a^2-b^2=(a+b)(a-b)$ with denominator $x^3$ used for $a-b$ and $x$ for $a+b$. Nov 3, 2022 at 4:44

Using $$\tan(x)=x+\frac{x^3}{3}+o(x^3)\quad \text{and}\quad \frac{1}{1+x}=1-x+o(x),$$ yields \begin{align*} \left(\cot(x)-\frac{1}{x}\right)\left(\cot(x)+\frac{1}{x}\right)&=\cot^2(x)-\frac{1}{x^2}\\ &=\frac{1}{\tan^2(x)}-\frac{1}{x^2}\\ &=\frac{1}{x^2+\frac{2}{3}x^4+o(x^4)}-\frac{1}{x^2}\\ &=\frac{1}{x^2}\left(\frac{1}{1+\frac{2}{3}x^2+o(x^2)}-1\right)\\ &=\frac{1}{x^2}\left(1-\frac{2}{3}x^2+1+o(x^2)\right)\\ &=-\frac{2}{3}+o(1). \end{align*}

$$\cot^2x-\frac1{x^2}=\csc^2x-1-\frac1{x^2}=\frac1{\sin^2x}-\frac1{x^2}-1=\frac{x^2-\sin^2x}{x^2\sin^2x}-1=\frac{(x-\sin x)(x+\sin x)}{x^4\frac{\sin^2x}{x^2}}-1$$

In the numerator, using $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$, thus,

$$\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}+...)(2x-\frac{x^3}{3!}+...)}{x^4\frac{\sin^2x}{x^2}}-1$$

Using $$\lim_{x\to0}\frac{\sin x}{x}=1$$, we get,

$$\lim_{x\to0}\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}+...)(2x-\frac{x^3}{3!}+...)}{x^4}-1\\=\lim_{x\to0}\frac{(\frac1{3!}-\frac{x^2}{5!}+..)(2-\frac{x^2}{2!}+...)}1-1\\=\frac26-1=-\frac23$$

Without using the derivatives, or Taylor or o(little) or O(big),$$\lim \limits_{x \to 0}\left( \cot^2 x-{1 \over x^2}\right) = \lim \limits_{x \to 0} \frac{x^2\cos^2 x-\sin^2 x}{x^2\sin^2 x} =$$$$\lim \limits_{x \to 0} \frac{x\cos x+\sin x}{x}\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3}$$$$\cdot\lim \limits_{x \to 0}\frac{x^2}{\sin^2 x}=\lim \limits_{x \to 0} \left( \cos x +\frac{\sin x}{x}\right)\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3}$$$$= 2 \cdot \lim \limits_{x \to 0}\frac{\cos x-x\sin x- \cos x}{3x^2}=-\frac 23$$