Prove $\frac{x}{1+n^2x^2}$ is uniformly convergent

Question

I need to prove that

\begin{equation}f_n(x)=\frac{x}{1+n^2x^2}\end{equation}

Converges uniformly to $0$. I've tried a solution:

(Scratchwork) Want $|f_n(x)-0|=|f_n(x)|=|\frac{x}{1+n^2x^2}|<\epsilon$, i.e. $\frac{|x|}{1+n^2x^2}<\epsilon$. Solving for $n$,

\begin{align*} \frac{|x|}{\epsilon}<1+n^2x^2 \\ \frac{|x|}{\epsilon}-1<n^2x^2 \\ \frac{|x|}{x^2 \epsilon}-\frac{1}{x^2}<n^2 \end{align*}

So choosing $N$ so that $N^2>\frac{|x|}{x^2 \epsilon}-\frac{1}{x^2}$ completes the proof. Would this be enough to prove uniform convergence on $\mathbb{R}$?

Edit My error above was that I chose $N$ in terms of $x$ which is no good. Thanks to "user" for pointing this out and thanks to Orman for his genius observation!

Answer 1

From the inequality $(1-n|x|)^2\geq 0$ we conclude that $1+n^2x^2\geq 2n|x|$ or $|f_n(x)|\leq\frac{1}{2n}$ for every $x$.

Answer 2

the problem is that your lower bound for $N$ depends on the variable $x$. To prove uniform convergence the bound must be independent of $x$. You may have to consider two separate cases: when $x$ is "small" and when $x$ is "large".

Answer 3

It's enough to prove that if $\lambda _n = \sup \{ |f_n(x)| | x \in \mathbb{R} \}$, then $\lim _{n \rightarrow \infty} \lambda _n = 0$.

So you just have to find the maximum of $|f_n(x)|$ and see that the limit goes to $0$. For that just differentiate your function find its zeroes.