Uniform convergence of $\frac {nx}{1+nx^2}$ on intervals $(0,\infty)$, $(0,1)$, and $(1,\infty)$

Question

Show whether or not the convergence of $f_n=\frac {nx}{1+nx^2}$ is uniform on the intervals $(0,\infty)$, $(0,1)$, and $(1,\infty)$.

I started checking if I could ensure $|\frac {nx}{1+nx^2}|\lt\epsilon$ for an arbitrary $\epsilon$ on $(0,\infty)$.

$|\frac {nx}{1+nx^2}-\frac{1}{x}|\le\frac {nx}{1+nx^2}+\frac{1}{x}\lt\frac{nx}{nx^2}+\frac{1}{x}=\frac{1}{x}+\frac{1}{x}=\frac{2}{x}\lt\epsilon$. ($x\gt\frac{\epsilon}{2}$ to satisfy the inequality, regardless of $N$.)

So, on $(0,\infty)$, is it correct to say we may, for any $\epsilon$, choose $x$ to be $x\le\frac{\epsilon}{2}$, thereby disproving $f_n$ is uniformly convergent on $(0,\infty)$? If so, how does this change the argument for the next two intervals?

Answer 1

Your argument is not valid. In fact the very first inequality you used spoils everything! Do not change the difference to the sum.

Now $|\frac {nx} {1+nx^{2}}-\frac 1 x|=\frac 1 {x(1+nx^{2})}$. On $(1,\infty)$ this is less than or equal to $\frac 1 {n+1}$ so the given sequence tends to $\frac 1 x$ uniformly.

On the other two intervals the convergence is not uniform as seen by putting $x=\frac 1 n$.