# Thread: did i solved this limit correctly

1. ## did i solved this limit correctly

f(x) and g(x) are differentiable on 0
f(0)=g(0)=0
$
\lim _{x->0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x->0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2
$

because i can use (sin x)/x=1 here
is it ok??

2. Originally Posted by transgalactic
f(x) and g(x) are differentiable on 0
f(0)=g(0)=0
$
\lim _{x->0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x->0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2
$

because i can use (sin x)/x=1 here
is it ok??
No because $f(x)=g(x)$ is permitted by your conditions, and so if $f(x) \not \equiv 0$ then the limit in the middle is $0$

CB