f(x) and g(x) are differentiable on 0
f(0)=g(0)=0
$\displaystyle
\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??
f(x) and g(x) are differentiable on 0
f(0)=g(0)=0
$\displaystyle
\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??