I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$
Has someone a different approach to solve this limit?
Best Answer
$$ \lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)= \lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^2\sin^2x}= \lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}\frac{x^2}{\sin^2x} $$ Apply l'Hôpital or Taylor expansion (better) to the first fraction.
For the Taylor expansion, it's easier to do it in pieces: $$ \sin x = x-\frac{x^3}{6}+o(x^4),\quad \cos x = 1-\frac{x^2}{2}+o(x^4) $$ so $$ \sin x-x\cos x=x-\frac{x^3}{6}-x+\frac{x^3}{2}+o(x^4)=\frac{x^3}{3}+o(x^4) $$ while $$ \sin x+x\cos x=x-\frac{x^3}{6}+x-\frac{x^3}{2}+o(x^4)=2x+o(x^2) $$ so $$ \lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}= \lim_{x\to 0}\frac{\frac{1}{3}x^3+o(x^4)}{x^3}\frac{2x+o(x^2)}{x}= \lim_{x\to 0}\left(\frac{1}{3}+o(x)\right)(2+o(x))=\frac{2}{3}. $$