I have solved the following exercise and I would like to know if there are any mistakes. Thank you.
Consider the sequence of functions defined by $g_n(x)=\frac{x^n}{n}$.
(a) Show $(g_n)$ converges uniformly on $[0,1]$ and find $g=\lim g_n$. Show that $g$ is differentiable and compute $g'(x)$ for all $x\in [0,1]$.
(b) Now, show that $(g_n')$ converges on $[0,1]$. Is the convergence uniform? Set $h=\lim g_n'(x)$ and compare $h$ and $g'$. Are they the same?
My solution:
(a) Let $\varepsilon>0$: then $|g_n(x)-0|=|\frac{x^n}{n}|\leq\frac{1}{n}$ for $x\in [0,1]$ so it suffices to choose $N>\frac{1}{\varepsilon}$ to guarantee that $|g_n(x)-0|<\varepsilon$ for all $n\geq N$ and $x\in [0,1]$ so $(g_n)$ converges uniformly to $0\equiv g(x)$ which, being a constant function, is differentiable and has $g'(x)\equiv 0$ too.
(b) $g_n'(x)=x^{n-1}$ converges pointwise to $0$ for $x\in [0,1)$ and to $1$ for $x=1$, so $h(x)=\lim g'_n(x)=\begin{cases} 0 & \text{if $x\in [0,1)$}\\ 1 & \text{if $x=1$} \end{cases}\neq g'(x)$ and the convergence is not uniform: by the Continuous Limit Theorem) since $(g'_n)$ is a sequence of continuous functions, if it converged uniformly its limit $h$ would be continuous too and that is not the case.
Yes, that's right. It's well written too. However, I think it's clearer for (a) to let $\epsilon > 0$ and then choose $N > 1/\epsilon$ and then proceed as you have.