Find the limit without using L'Hopital's rule $lim_{x to 0} (x-tan x)/(x tan x)$.

I solved it with L'Hopital's rule but I want to find out how can I solve it without using L'Hopital's rule.

Answer

$$\lim_\limits{x\to 0}\dfrac{x-\tan x}{x\tan x}$$ $$=\lim_\limits{x\to 0}\bigg(\dfrac{x}{x\tan x}-\dfrac{\tan x}{x\tan x}\bigg)$$ $$=\lim_\limits{x\to 0}\bigg(\dfrac{1}{\tan x}-\dfrac{1}{x}\bigg)$$ $$=\lim_\limits{x\to 0}\bigg(\dfrac{\cos x}{\sin x}- \dfrac{1}{x}\bigg)$$ $$=\lim_\limits{x\to 0}\bigg(\dfrac{\cos x}{\dfrac{x\sin x}{x}}- \dfrac{1}{x}\bigg)$$ $$=\lim_\limits{x\to 0}\bigg(\dfrac{\cos x}{\dfrac{\sin x}{x}}\dfrac{1}{x}- \dfrac{1}{x}\bigg)$$ $$=\lim_\limits{x\to 0}\bigg(\dfrac{1}{x}-\dfrac{1}{x}\bigg)$$ $$\text{Because, $\lim_\limits{x\to 0}\cos x=1~~and~~\lim_\limits{x\to 0}\dfrac{\sin x}{x}=1$}$$ $$=0$$