Uniform convergence of $\lim_{n \to \infty} \frac{nx}{1 + n^2x^2} \text{ where } 0 \le x \le 1$

Question

Prepering to second exam in Calculus 2,

and I remember this question from the first exam that I fell on:

Let $f_n(x)$ be the function sequence $\{\frac{nx}{1 + n^2x^2}\}_{n=1}^\infty$

Let $f(x)$ be the limit function such that $\lim_{n \to \infty} \frac{nx}{1 + n^2x^2}$ convergences to.

Does $f_n(x)$ convergents uniformly to $f(x)$?

As for the answers, the answer is no.

Can you please explain me why?

Thanks in advance!

@DavidMitra The pointwise limit is $0$ (and this is the reason I though it convergents uniformly). $f_n(1/n)$ seems to be 1. Well, I've got the point. — Billie, Feb 06 '14 at 10:17
Yes, the pointwise limit is $0$ on all of $\Bbb R$. $f_n(1/n)$ is $1/2$. Now, if $(f_n)$ converged uniformly, there would be an $N$ so that for all $n\ge N$ and all $x$, we would have $|f_n(x)|<1/2$. Presents a quandry, no? — David Mitra, Feb 06 '14 at 10:19
See also: To prove that $f_n(x) = \frac{nx}{1+n^2x^2}$ does not uniformly converge to $f(x) = 0$ on $[0,1]$ — Martin Sleziak, Sep 19 '18 at 14:06

score 1 · Accepted Answer · answered Feb 06 '14 at 10:34

1

If you investigate the $$c_n=\max\mid f_n(x)-f(x)\mid=\max\limits_{x\in[0,1]}\mid \frac{nx}{1+n^2x^2}\mid$$ then you find that it takes its maximum at $x=\frac{1}{n}$ so $$c_n=\frac{n.\frac{1}{n}}{1+n^2\left(\frac{1}{n}\right)^2}=\frac{1}{2} $$which means that $f_n(x)$ converges nonuniformly.

answered Feb 06 '14 at 10:34

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How do you conclude that "its maximum at $x=\frac{1}{n}$"? – Lior Jul 11 '23 at 15:17

Uniform convergence of $\lim_{n \to \infty} \frac{nx}{1 + n^2x^2} \text{ where } 0 \le x \le 1$

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