Pointwise limit,$f$, of the sequence is not bounded

Question

Question: Let $f_n(x) := \frac{nx}{1+nx^2}$ for $x \in A := [0, \infty)$. Show that each $f_n $is bounded on $A$, but the point-wise limit of $f$ of the sequence is not bounded on $A$.

Does $(f_n)$ converge uniformly to $f$ on $A$?

I'm having trouble showing the first part of the question, let alone completing it.

I know that $f_n(x)\leq \frac{1}{x},$ but that doesn't seem to getting me anywhere.

Answer 1

Each $f_n$ is continuous, hence bounded on $[0,1]$, say $0\le f_n(x)\le M_n$. Together with your result $f_n(x)\le \frac1x$ (for $x>0$), we conclude $0\le f_n(x)\le \max\{M_n,1\}$.

It should be clear that $f(0)=0$ and $f(x)=\frac1x$ for $x>0$, which is not bounded (if $M>0$ then $f(\frac1{M+1})=M+1>M$).

You should know that the limit of a uniformly convergent sequence of continuous functions is continuous; $f$ is not.

Answer 2

1.) If $x=0$ then for all $n\in\mathbb{N}$ we have $f_n(0)=0$. If $x\in(0,\infty)$ we have that $1\leq\frac{1}{x}+nx\implies \frac{1}{\frac{1}{nx}+x}=\frac{nx}{1+nx^2}\leq n$. So each $f_n(x)$ is bounded by $n$.

2.) Now, consider $\lim\limits_{n\to\infty}f_n(x)=\lim\limits_{n\to\infty}\frac{nx}{1+nx^2}=?$. If we plug in $x$ values in a "certain manner" we see that the limit function goes to $\infty$. Let's plug in $x$-values according to the sequence $x_n:=\frac{1}{\sqrt{n}}$, clearly $x_n\in(0,\infty)$ for all $n\in\mathbb{N}$. Then we see that $f_n(x_n)=\frac{nx_n}{1+nx_n^2}=\frac{\sqrt{n}}{2}$ and hence $\lim\limits_{n\to\infty}f_n(x_n)=\infty$. So the limit function $\lim\limits_{n\to\infty}f_n(x)$ is unbounded and therefore can't be uniformous convergent.