I'm asked to determine if the following sequences converges and how:
$$f_n= \frac{nxe^{-nx}}{n^2x^2+1}$$
It's easy to see that $\lim_{n \to \infty} f_n=0$ and so $f_n$ converges pointwise to $f=0$. To see if the convergence is also uniform I have to evaluate
$$\lim_{n \to \infty}\sup_{\mathbb{R}}||f_n(x)-f(x)||= \lim_{n \to \infty}\sup_{\mathbb{R}}||\frac{nxe^{-nx}}{n^2x^2+1}||$$
so I tried to study $f_n'(x)$ but this lead nowhere (I suppose I don't have to use the algebraic formula for finding roots of polynomials of degree $3$):
$$f_n'(x)=0 \iff \frac{(ne^{-nx}-n^2xe^{-nx})(n^2x^2+1)-2n^2x(nxe^{-nx})}{(n^2x^2+1)^2}=0 \iff n^3x^3+n^2x^2+nx+1=0$$
and I'm stuck.
Can I have a help, please?
EDIT
I'm asked if and where the convergence is uniform.
Best Answer
A mistake In your third degree equation $$n^3x^3+n^2x^2+nx-1=0$$
here is an other way;
$(f_n)$ converges pointwise to zero at $[0,+\infty)$.
for $x<0$, $$\lim_{n\to+\infty}f_n(x)=-\infty$$
on the other hand, for $n>0$
$$|f_n(\frac 1n)-0|=\frac{1}{2e}$$
thus $$\sup_{x\ge 0}|f_n(x)-0|\ge \frac{1}{2e}$$ the convergence is not uniform.