# Uniform convergence of $f_n = \frac{nx}{1+nx^2}$.

let $$f_n(x)$$ = $$\frac{nx}{1+nx^2}$$ for $$x\ge 0$$,

(a) show that $$\{f_n\}$$ converges uniformly on $$[M,\infty]$$ for any $$M>0$$

(b) show that $$\{f_n\}$$ does not converge uniformly on $$[0,\infty)$$.

I tried solving this problem by using Weierstrass $$M_n$$ test for sequence of functions:

$$f_n(x) \to f(x)=\frac{1}{x}.$$

Now, I am stuck in finding

$$M_n=\sup\{\left\lvert f_n(x)-f(x)\right\rvert\},$$ where the supremum is taken over $$[M,\infty)$$ for (a) or $$[0,\infty)$$ for (b).

Also, is it true in general that if $$M_n$$ does not tend to $$0$$, then $$\{f_n\}$$ does not uniformly converge to $$f$$?

Hint 1: For $|x|\ge M$, \begin{align} \left|\frac{nx}{1+nx^2}-\frac1x\right| &=\frac1{x(1+nx^2)}\\ &\le\frac1{nM^3} \end{align} Hint 2: The Uniform Limit Theorem.

Instead of using the uniform limit theorem, we show directly that $$f_n$$ does not converge uniformly on $$(0,M)$$ for all $$M>0$$. Clearly the pointwise limit is

$$g(x) = \begin{cases} \frac{1}{x} & x>0, \\ 0 & x =0 \end{cases}.$$

Then when $$x>0$$,

$$|f_n(x) - g(x) | =\frac{1}{x(1+nx^2)}.$$

And this implies

$$\left|f_n\left(\frac{1}{\sqrt{n}}\right) - g\left(\frac{1}{\sqrt{n}}\right)\right| = \frac{\sqrt n}{2} \ge \frac{1}{2}$$ for all $$n$$. Then for any $$M>0$$,

$$\sup_{x\in (0,M)}|f_n(x) - g(x)| \ge \frac 12$$ for all $$n\ge 1/M^2$$. Thus $$\{f_n\}$$ does not converge uniformly to $$g$$ on $$(0,M)$$ for any $$M>0$$.