Choice of $N$ for $\frac{nx}{1+nx^2}$

Question

Let $$f_x(x) = \frac{nx}{1+nx^2}$$

Is the convergence uniform on $(0, \infty)$?

We know that it's pointwise convergent onf $f(x)= \frac{1}{x}$ as we take the limit $\lim_{n \to \infty} f_n(x)$.

However, to show that it's not uniform we proceed with the following:

$$|f_n(x) - f(x)| = \left| \frac{nx}{1+nx^2} - \frac{1}{x} \right| = \frac{1}{x+nx^3}$$

What throws me off next is why we choose the following $N$

$$N \ge \frac{1-\epsilon x}{\epsilon x^3}$$

to make $|f_n(x) - f(x)| < \epsilon$

Answer 1

Referring to what "throws you off":

$$N \ge \frac{1-\epsilon x}{\epsilon x^3}$$

This $N$ is not chosen. The idea here is to show that such an $N$ does not exist. This then means that the convergence cannot be uniform.

Here the complete proof by contradiction:

Assume that the convergence is uniform and let $\epsilon >0$ be arbitrary but fixed. Then, there should be an $N\in\mathbb{N}$, such that for all $x\in (0,\infty)$

$$\frac 1{x+Nx^3}\leq \epsilon$$

Solving this inequality vor $N$ gives

$$N \ge \frac{1-\epsilon x}{\epsilon x^3}$$

This means, this $N$ should satisfy above inequality for all $x\in (0,\infty)$. Now, choosing $0<x$ small enough such that $2\epsilon x < 1$ you get

$$N \ge \frac{1-\epsilon x}{\epsilon x^3}> \frac{2\epsilon x-\epsilon x}{\epsilon x^3} = \frac 1{x^2} \stackrel{x\to 0^+}{\longrightarrow} \infty$$

This shows that such an $N$ does not exist. Hence, the convergence cannot be uniform.

Answer 2

The convergence is not uniform. You have$$f_n'(x)=\frac{n-n^2 x^2}{\left(n x^2+1\right)^2}$$and it is easy to deduce from this that$$\max|f_n|=f_n\left(\frac1{\sqrt n}\right)=\frac{\sqrt n}2.$$So, since $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function, the convergence is not uniform.