uniform convergence fn to f

Question

I'm given the following function sequence:

$$f_n = \frac{nx}{1+nx^2}, \forall x \in A = [0,\infty].$$

I show the following that:

$$\lim_{n \to \infty} \frac{nx}{1+nx^2} \le \frac{nx}{nx^2} \le \frac{1}{x}.$$

And thus my convergent function I compute is $f(x) = \frac{1}{x}.$

However the answer appears to be $f(x) = \frac{1}{2x}$ using A/G mean inequality.

This leads to my next question that if $f_n \to f$ converges EITHER point wise or uniformly to $f$, is $f$ unique?

Answer 1

The question of uniform convergence is solved by considering the supremum of the fraction $$\sup\limits_{(0,+\infty)}\left|\frac{nx}{1+nx^2}-\frac{1}{x}\right| = \sup\limits_{(0,+\infty)}\frac{1}{x(1+nx^2)}=+\infty$$

Answer 2

The pointwise convergent function $f$ is not $f = \frac{1}{x}$ on the domain given A. Please note the correct answer below for f:

$f = 0: x = 0$

$f = \frac{1}{x}: x > 0 $

Note in the sequence $f_n(0) = 0, \forall n \in \mathbb N$

Convergence is not uniform since f is not continuous.

If $f_n \to f$ converges pointwise f is unique and which can be shown by a contradiction taking $2\epsilon = |f_1 - f_2|$ where $f_1 \ and \ f_2$ are diffrent pointwise convergence functions and resulting in a $ 2\epsilon < 2\epsilon \ \forall \epsilon > 0$ which statement is a contradiction.