Uniform convergence of $f_n(x)= nx/(n^2x^2+1)$

Question

I am trying to find whether $f_n(x)= nx/(n^2x^2+1)$ is uniformly convergent or not.

I found that the pointwise limit was $0$ except for $x=1/n$. Can I then say that $f$ is uniformly convergent on $(-\infty,-a) \cup (a,\infty)$ for all $a>0$ so that, for $n$ large enough, I cannot take $1/n$ because it would be less than $a$, and then I always have that $|f_n(x)-f(x)| < \epsilon$ ?

Thank you!

Answer 1

Asserting that the pointwise limit of a sequence $(f_n)_{n\in\Bbb N}$ of functions from $\Bbb R$ into $\Bbb R$ is some $f(x)$ unless $x=\frac1n$ makes no sense, since $\frac1n$ is not a fixed number. Besides, for every real number $x$ you do have $\lim_{n\to\infty}\frac{nx}{n^2x^2+1}=0$. So, your sequence converges pointwise to the null function.

However, since $(\forall n\in\Bbb N):f_n\left(\frac1n\right)=\frac12$, you know that $\sup_{x\in\Bbb R}\bigl|f_n(x)-0\bigr|\geqslant\frac12$, and therefore the convergence is not uniform. Otherwise, you would have $\lim_{n\to\infty}\sup_{x\in\Bbb R}\bigl|f_n(x)-0\bigr|=0$.