Show that the convergence cannot be uniform on $[0, \infty)$ for $$g_n(x) = \frac{x}{1+x^n}$$
Here's what I have tried:
We know it's pointwise convergent on $g(x) = x$ on the limit $[0, \infty)$.
$$|g_n(x)-g(x)| = \left|\frac{x}{1+x^n} - x \right| = \frac{x^n}{1+x^n}$$
To make $|g_n(x) - g(x)| < \epsilon$ we must find an $N$
$$\epsilon \ge \frac{x^N}{1+x^N} $$
By re-arranging for $N$ we have $$N \ge \frac{\log(-\epsilon)-\log(\epsilon-1)}{\log(x)}$$
Then I want to restrict $x$ such that $x \ge 0$ so that the numerator goes away and I only get $\lim_{x\to\infty}\frac{1}{\log(x)}=0$ to show it's not uniform. However, I'm not sure how I can show this?
Actually, your sequence of functions converges pointwise to$$\begin{array}{rccc}g\colon&(0,\infty)&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\in(0,1)\\\frac12&\text{ if }x=1\\0&\text{ if }x>1.\end{cases}\end{array}$$Since each $g_n$ is continuous but $g$ isn't, the convergence is not uniform.
If you want to prove that the convergence is not uniform using the definition of uniform convergence, that can be done too. Take $\varepsilon=\frac16$. Let $N\in\Bbb N$. Take $x=\sqrt[N]{1/2}$. Then\begin{align}\left|g(x)-g_N(x)\right|&=\left|\sqrt[N]{\frac12}-\frac{\sqrt[N]{1/2}}{1+1/2}\right|\\&=\frac13\sqrt[N]{\frac12}\\&\geqslant\frac13\times\frac12\\&=\frac16\\&=\varepsilon.\end{align}