How to approach this exercise? I've tried a lot of methods and always reached 0/0.

\lim_{x \to 0} \frac{\sin x - \tg x}{\sin^2 x \tg x }

How to approach this exercise? I've tried a lot of methods and always reached 0/0.

\lim_{x \to 0} \frac{\sin x - \tg x}{\sin^2 x \tg x }

lemon14: please check your LaTex notation - what operation are you trying to signify using "\tg"?

I think you mean
\lim_{x\rightarrow 0} \frac{\sin x-\tan x}{\sin^2 x\tan x}.

The case 0/0 is very common in limits, so you have essentially to simplify something in order
to get a value (for instance, even \lim_{x\rightarrow 0} \frac{x^2}{x} is a 0/0
but you can simplify it). A first rule you can use is de l'Hopital rule: essentially
(under rather weak conditions) if

\lim_{x\rightarrow 0} \frac{f(x)}{g(x)}
is a 0/0, then
\lim_{x\rightarrow 0} \frac{f(x)}{g(x)}=\lim_{x\rightarrow 0} \frac{f'(x)}{g'(x)}.

An alternative approach it to use several identities: first \sin x=\cos x\tan x so
\frac{\sin x-\tan x}{\sin^2 x\tan x}=\frac{\sin x\cos x-\sin x}{\sin^3 x}=
\frac{\cos x-1}{\sin^2 x}
Using \lim_{x\rightarrow 0} \frac{\sin x}{x}=1
you get
\lim_{x\rightarrow 0}\frac{\cos x-1}{\sin^2 x} =\lim_{x\rightarrow 0}\frac{\cos x-1}{ x^2}=-1/2 .