Prove $x/(1+nx^2)$ uniformly converges to zero.

Question

$f_n(x) = \frac{x}{1+nx^2}$. Prove $f_n$ converges pointwise over all $\mathbb{R}$ to $f(x) = 0$. Is this convergence uniform?

My proof is below.

1. Can you verify it?
2. Is there a simpler or more direct approach than taking $1/f_n$?

Proof: Choose $N > 1/\epsilon|x|$. Then $|\frac{x}{1+nx^2}| < |\frac{x}{nx^2}| = |\frac{1}{nx}| < \epsilon$.

The convergence is uniform. Choose $N > 1/\epsilon^2$. For $x=0$, the matter is trivial. Else, $1/f_n(x) = \frac{1}{x} + nx$. Assume $x>0$. If $x \leq 1/\sqrt{n}$, then $1/x \geq \sqrt{n}$, and if $x \geq 1/\sqrt{n}$, then $nx \geq 1/\sqrt{n}$. Hence, $1/f_n(x) \geq 1/\sqrt{n}$, and $|f_n(x)| < \epsilon$ for all $n > N$. A similar argument holds for $ x < 0$.

Answer 1

As regards your proof, I did not get the last part: "Hence, $1/f_n(x) \geq 1/\sqrt{n}$, and $|f_n(x)| < \epsilon$ for all $n > N$."

Another (simpler) way: for any $x\in \mathbb{R},$ $$(1-\sqrt{n}x)^2\geq 0\implies 1+nx^2\geq 2\sqrt{n}x\implies |f_n(x)|\leq \frac{1}{2\sqrt{n}}.$$ Then uniform convergence of $(f_n)_n$ over $\mathbb{R}$ follows instantly.