Is it possible to find $$\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}$$ without l'Hospital's rule?
I have $\lim_{x\to 0}\frac{\tan x}{x}=1$ proved without H. but it doesn't help me with this complicated limit (however I'm sure I have to use it somehow).
I know the answer is $\frac{1}{3}$, so I tried to estimate: $0<\frac{\tan x-x}{x^2\tan x}\le\frac{1}{3}\cdot\frac{\tan x}{x}+g(x)$ with various $g$ and prove that $g(x)\to 0$, but with no result.
Best Answer
$$L=\lim_{x\to 0}\frac{x-\tan x}{x^2 \tan x}=\lim_{x\to 0}\frac{\cos x-\frac{\sin x}{x}}{x^2}\cdot\frac{x}{\sin x}=\lim_{x\to 0}\frac{1-2\sin^2\frac{x}{2}-\cos\frac{x}{2}\frac{\sin(x/2)}{(x/2)}}{x^2} $$ gives $L=A+B$ where: $$ A = -\frac{1}{2}+\lim_{x\to 0}\frac{1-\cos\frac{x}{2}}{x^2} = -\frac{1}{2}+\lim_{x\to 0}\frac{2\sin^2\frac{x}{4}}{x^2} = -\frac{1}{2}+\frac{1}{8}=-\frac{3}{8}$$ and $$ B = \lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{\sin(x/2)}{x/2}\right)=\frac{1}{4}\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{\sin x}{x}\right)$$ (assuming such a limit exists) fulfills: $$ 3B = 4B-B = \lim_{x\to 0}\left(\frac{\sin(x/2)}{x/2}-\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{2\sin(x/2)}{\sin (x)}-1\right)$$ so that: $$ B = \frac{1}{3}\lim_{x\to 0}\frac{1}{x^2}\left(\frac{1}{\cos\frac{x}{2}}-1\right)=\frac{1}{3}\lim_{x\to 0}\frac{1-\cos\frac{x}{2}}{x^2}=\frac{1}{3}\lim_{x\to 0}\frac{2\sin^2\frac{x}{4}}{x^2}=\frac{1}{24}$$ and $L=A+B=-\frac{3}{8}+\frac{1}{24}=\color{red}{\large-\frac{1}{3}}$.