1. ## Question about uniform convergence

Hi, so Im studying for finals and am stuck on a problem. I got referred to this forum from a friend, and I had a quick question about uniform convergence.
we have a function which is:

$f_n(x)\frac{nx}{nx+1}$ for all n greater than or equal to 1.

Is this uniformly convergent at [0,1]?
Then if we were to fix r>0, is it uniformly convergent at [r, infinity]?

Also, there was another part of the problem. the professor wrote something like
$f(x) = \lim_{x\to\infty}f(x^n)$
He's a messy writer, and i do believe that was what he wrote, but Im not too sure if that fits into the problem at all, or if it even does.

Any help is appreciated. Thanks!

2. Originally Posted by wontonsoup
$f_n(x)\frac{nx}{nx+1}$ for all n greater than or equal to 1. Is this uniformly convergent at [0,1]?
Every $f_n$ is continuous on $[0,1]$ , however:

$f(x)=\displaystyle\lim_{n \to \infty}f_n(x)=\begin{Bmatrix}0&\textrm{if}&x=0\\1& \textrm{if}&x\in(0,1]\end{matrix}$

is not continuous on $[0,1]$, so by a well known theorem, the convergence is not uniform.

Then if we were to fix r>0, is it uniformly convergent at [r, infinity]?
Yes, now

$f(x)=\displaystyle\lim_{n \to \infty}f_n(x)=1$

Prove that for every $\epsilon>0$ there exists $n_0\in\mathbb{N}$ such that

$|f_n(x)-f(x)|<\epsilon$ for $n\geq n_0$ and for all $x\in[r,+\infty)$

Also, there was another part of the problem. the professor wrote something like
$f(x) = \lim_{x\to\infty}f(x^n)$
He's a messy writer, and i do believe that was what he wrote, but Im not too sure if that fits into the problem at all, or if it even does.
Possibly is:

$f(x)=\displaystyle\lim_{n \to \infty}f_n(x)$

Fernando Revilla