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lemon14 · Jan 12, 2012, 05:58 AM

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

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

ebaines · Jan 12, 2012, 06:37 AM

Originally Posted by lemon14

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"?

Aurora2000 · Jan 12, 2012, 07:04 AM

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
$\frac{\sin x-\tan x}{\sin^2 x\tan x}=\frac{\sin x\cos x-\sin x}{\sin^3 x}=<br /> \frac{\cos x-1}{\sin^2 x}$
Using $\lim_{x\rightarrow 0} \frac{\sin x}{x}=1$
you get
$\lim_{x\rightarrow 0}\frac{\cos x-1}{\sin^2 x} =\lim_{x\rightarrow 0}\frac{\cos x-1}{ x^2}=-1/2$ .