Exercise Let $f_n(x) = \frac{x}{1+n^2x^2}$ for each $n \in \mathbb{N}$ and $x \in (0, \infty)$ Show that $\{f_n\}$ converges uniformly.
Proof Attempt
To show uniform convergence, we must show
$$\forall \varepsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, |f_n(x) - f(x)| < \varepsilon \, \forall x \in (0, \infty)$$
Let $x > 0$ and $f_n(x) = \frac{nx}{1+n^2x^2}$. Observe that
$$f(x) = \displaystyle\lim_{n \to \infty} \frac{x}{1+n^2x^2} = 0$$
Now we consider the following:
$$ |f_n(x) -f(x)| = \Big| \frac{x}{1+n^2x^2} - 0 \Big| = \Big| \frac{x}{1+n^2x^2} \Big| \leq ?$$
Would it be beneficial here to invoke the triangle inequality or is it better to construct a compound inequality and show uniform convergence via the squeeze theorem for limits? If the triangle inequality is to be used, what value would be helpful to add and subtract? If the squeeze theorem is better, what interval would be appropriate to use? Any hints here would be greatly appreciated.
Notice that, whenever $x\in(0,+\infty)$, the next result holds: \begin{align*} 1 + n^{2}x^{2} \geq 2\sqrt{n^{2}x^{2}} = 2nx \Rightarrow \frac{x}{1 + n^{2}x^{2}} \leq \frac{1}{2n} \end{align*}
Moreover, remember as well that uniform convergence is the same thing as convergence according to the sup metric. Having said that, let $n\geq n_{\varepsilon}$. Then we have: \begin{align*} \sup_{x\in(0,\infty)}\left|\frac{x}{1 + n^{2}x^{2}}\right| = \frac{1}{2n} \leq \frac{1}{2n_{\varepsilon}} := \varepsilon \end{align*}
Can you take it from here?