the exercise asks me to prove that the limit $ \lim_{x \to \infty}(\frac{-x^{2n-2}}{(2n+4)(1+x^2)^{n+2}})= \frac{n-1}{n+2} $
bag out of bounds $ \frac{1}{(2n+4)}\lim_{x \to \infty}(\frac{-x^{2n-2}}{(1+x^2)^{n+2}}) $ then I apply L'hopital's rule and derive the expression and i get this
$\frac{1}{(2n+4)}\lim_{x \to \infty}(\frac{-2n-2*x^{2n-3}*(1+x^2)^{n+2}-[(8n+2)*(1+x^2)^{n+1}*2x*x^{2n-2}]}{((1+x^2)^{n+2})^2})$
If I simplify the previous expression I get
$\frac{1}{(2n+4)}\lim_{x \to \infty}(\frac{-2x^{-3+2n}(-3x^2+n-1)}{(x^2+1)^{n+3}})$
but that limit is 0 I do not know if I did something wrong o I need to change something
