Sequence of functions convergence investigation and question

Question

So i have a sequence of functions defined as:

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

So i have to investigate convergence.

So first i found limit function:

$$F(x)=0$$

So that's for pointwise convergence. (Is it enough?)

Then i concluded this: $$f_n(x) \in [-\frac{1}{2},\frac{1}{2}]$$ Where $f_n(\frac{1}{n})=\frac{1}{2}$

Hence

$$|f_n(\frac{1}{n})-f(\frac{1}{n})|=\frac{1}{2}$$

So its not uniformly convergent. But i have an additional question, since i think it could be uniformly convergent on $[a,\infty)$ for a>0. But i want to prove this, i don't know whether i should try the same method as for whole $\mathbb{R}$ or is there a trick for it.

Any help would be appreciated.

Answer 1

Using your work, $f$ does not converge uniformly on $\mathbb{R}$ or any interval $I$ including $x = 0$ since for all sufficiently large $n$ we have $1/n \in I$ and

$$\sup_{x \in I} |f_n(x)- 0| = \sup_{x \in I} \frac{n|x|}{1 + n^2 x^2}= \frac{1}{2}$$

However, convergence is uniform on $S =(-\infty,-a] \cup [a, \infty)$ for any $a > 0$ since for all $n > 1/a$ we have

$$\sup_{x \in S} |f_n(x)- 0| = \sup_{x \in S} \frac{n|x|}{1 + n^2 x^2}= \frac{na}{1 + n^2 a^2} \to 0$$