Uniform convergence of sequence of function $f_n(x) = \frac{nx^4+1}{nx^4+2x+3}e^{-nx^2}$ on the interval $(1,+ \infty)$ I need to prove that the sequence of functions $f_n(x)$ is uniformly convergent to $f$ on the interval $(1,+ \infty)$.
I've already shown that  $f_n(x) = \frac{nx^4+1}{nx^4+2x+3}e^{-nx^2}$  is pointwise convergent to $0$ on $(1,+ \infty)$.
It remains to show that: $\:$ $\underset{x\in (1,+ \infty)}{sup}|f_n(x)-f(x)| \rightarrow 0$ for n $\rightarrow + \infty $
$sup|f_n(x)-f(x)|= sup|\frac{nx^4+1}{nx^4+2x+3}e^{-nx^2}|=sup\frac{x^4+1}{nx^4+2x+3}e^{-nx^2}\\$.
I tried to compute the derivate $f_n'(x)=\frac{4nx^3(nx^4+2x+3)-(nx^4+1)(4nx^3+2)}{(nx^4+2x+3)^2}e^{-nx^2}+\frac{nx^4+1}{nx^4+2x+3}*(-2nxe^{-nx^2})$ 
but i dont' find the global maximum point of $f_n$.
How can i find this point to check the uniform convergence ?
Thanks for the help in advance!!

Notice that for all $x>1$
$$\sup_{(1,\infty)} \frac{nx^4+1}{nx^4+2x+3}e^{-nx^2}\leq \sup_{(1,\infty)} \frac{nx^4+1}{nx^4}e^{-nx^2}$$
because dividing by a larger number gets a smaller number. So now we want to show
$$\lim_{n\rightarrow\infty} \sup_{(1,\infty)} \Big[\Big(1+\frac{1}{nx^4}\Big)e^{-nx^2}\Big]=0.$$
Since $1/(x+\delta) < 1/x$ and $e^{-x}<e^{-(x+\delta)}$ for all $\delta>0$, then the supremum is at $x=1$. Then it is sufficient to show that
$$\lim_{n\rightarrow\infty} \Big(1+\frac{1}{n}\Big)e^{-n}=0.$$
