Limits: $\lim_{n\rightarrow\infty}\frac{nx}{1+n^2x^2}$ on $I=[0,1]$

Question

I'm doing an assignment for my analysis course on the uniform convergence. And I have to assess the uniform convergence of the sequence $$f_n(x)=\frac{nx}{1+n^2x^2}$$ on the intervals $I=[1,2]$ and $I=[0,1]$. Before assessing whether there is uniform convergence, we need to provide the pointwise limit of the sequences. Now, for the first interval everything is okay, but I'm running into issues with finding the pointwise limit for the second interval.

It's clear to me that for $0<x\leqslant1$ the limit of the sequence should be zero. But when $x=0$ I'm not sure if this is also the case. Clearly filling in $x=0$ for any finite $n$ would give zero for $f_n(0)=0$, but how can we be sure this is still the case when $n\rightarrow\infty$?

Also, for the uniform convergence part of the assignment, I've gotten as far as seeing that I should use the fact that $f_n(\frac{1}{n})=\frac{1}{2}$, but I'm not really sure yet how to use this.

Any help is much appreciated!

Answer 1

You have proved a pointwise convergence to $0$, that is for any $x \in [0,1]$, as $n \to \infty$ $$ f_n(x)=\frac{nx}{1+n^2x^2} \to 0. $$ There is no uniform convergence over $[0,1]$ since $$ \sup_{x \in [0,1]}\left|f_n(x)-0\right|=\left|f_n\left(\frac1n\right)\right|=\frac12 $$ and this does not tend to $0$.

Answer 2

Let's show that $f_n(x)$ doesn't converge uniformly to zero using negation of definition $$\exists \varepsilon = \frac{1}{4}\forall n^*\in\mathbb{N}\exists n=n^*\exists x=\frac{1}{n}\in I \implies \Big|\frac{nx}{1+n^2x^2}\Big| = \Big|\frac{n\frac{1}{n}}{1+n^2\frac{1}{n^2}}\Big|=\frac{1}{2} > \frac{1}{4} $$

Answer 3

For $x\in [1,2]$ we have

$$0 < f_n(x) = \frac{nx}{1+n^2x^2} < \frac{nx}{n^2x^2} = \frac{1}{nx} \le \frac{1}{n}.$$

This shows $f_n \to 0$ uniformly on $[1,2].$ (In fact $f_n \to 0$ on any $[a,\infty),$ where $a>0.$)