Consider the sequence of functions $f_n(x)=\frac{x}{nx+1}, x\in(0,1).$ Does it converge uniformly?
If $f_n$ are differentiable $\forall n$ and $f_n(x)$ converges pointwise to $f(x)$, this technique is often used. I' ve tried to apply it:
Pointwise convergence: if $x\in(0,1), \lim_{x \rightarrow \infty} \frac{x}{nx+1} = 0.$
Uniform convergence: $f_n$ are differentiable and $f_n$ converges pointwise to $0$, so we now proceed to find $\sup_{0\leq x\leq 1} |f_n(x)-f(x)|$. In order to find it, we maximize $|f_n(x)-f(x)|=|\frac{x}{nx+1}|$;
$(\frac{x}{nx+1})' =\frac{nx+1-nx}{(nx+1)^2}=\frac{1}{(nx+1)^2}=0$. There is no $x\in(0,1)$ such that $\frac{1}{(nx+1)^2}=0$ so $|f_n(x)-0|$ doesn' t reach it's maximum for $x \in (0,1)$. My question is, is this enough to conclude that $f_n$ does not converge to $f$ uniformly? And if so, why?
Your sequence converges uniformly to the null function. Note that each $f_n$ is increasing and that therefore$$(\forall n\in\mathbb{N})(\forall x\in(0,1)):f_n(x)\leqslant\lim_{x\to1}f_n(x)=\frac1{n+1}$$and that $\lim_{n\to\infty}\frac1{n+1}=0$.