Uniform Continuity of a Sequence of Functions with a Piece-Wise Defined Limit

Consider the sequence of functions $$h_n(x)=\frac{x}{1+x^n}$$ over the domain $[0,\infty)$.

I found its pointwise limit to be

$$h(x)=\begin{cases} x & \text{ if } 0\leq x<1\text{,} \\ 1/2 & \text{ if } x=1\text{,} \\ 0 & \text{ if } x>1\text{.} \end{cases}$$

But I have a "gut-feeling" that the function may not be uniformly continuous over that interval (because it is piece-wise defined), so I am now trying to find a smaller set for which it is:

I naturally went with the choice of $(0,1)$, but then $$\left|h_n(x)-h(x)\right|=\left|\frac{x}{1+x^n}-x\right|=\frac{x^{n+1}}{1+x^n}<x<\epsilon,$$

which does not provide the necessary information I need to conclude that the function is uniformly continuous over that interval.

Do you guys have any ideas? Thanks!

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$h$ is not continuous, so in particular this function cannot be uniformly continuous. – Davide Giraudo Mar 7 '12 at 16:00
But could it be shown that it is uniformly continuous over a smaller set? – EsX_Raptor Mar 7 '12 at 16:01
$h$ is continuous on $[0,1)$ and $(1,+\infty)$ (uniformly in the second set). What is exactly the question? – Davide Giraudo Mar 7 '12 at 16:03
I just needed to find a set over which the sequence converges uniformly. :) I will try the set $(1, \infty)$, as you suggested. – EsX_Raptor Mar 7 '12 at 16:05
isn't $h$ also uniformly continuous on $[0,1)$. Pf: let $\varepsilon=\delta$ then $|f(x)-f(y)|=|x-y|<\delta=\varepsilon$ – Galois Mar 7 '12 at 16:06

Hints:

As noted in the comments, $(h_n)$ does not converge uniformly on $[1,\infty)$ or on $[0,1]$, since its limit function is discontinuous on those intervals.

By considering the values $x_n=1-{1\over n}$, you should be able to show that $(h_n)$ does not converge uniformly on the interval $[0,1)$ (compute $\lim\limits_{n\rightarrow\infty} h_n(x_n)$). By considering $x_n=1+{1\over n}$, you should be able to show that $(h_n)$ does not converge uniformly on the interval $(1,\infty)$ (note this would also verify the results of the preceeding paragraph).

But, consider what type of convergence you have on a set of the form $[0,1-\epsilon]$ or $[1+\epsilon,\infty)$ for a fixed $\epsilon>0$. Here, use the inequalities $$\Bigl| {x\over 1+x^n} -x \Bigr|=\Bigl| {x^{n+1}\over 1+x^n}\Bigr|\le x^{n+1},$$ for the $[0,1-\epsilon]$ case; and $$\Bigl| {x\over 1+x^n} \Bigr|\le\Bigl| {x \over x^n}\Bigr|={1\over x^{n-1}},$$ for the $[1+\epsilon,\infty)$ case.

Looking at the graphs of the $h_n$ should help:

Above are plotted $\color{darkred}{h_2}$, $\color{darkslateblue}{h_4}$, $\color{olivedrab}{h_{10}}$, $\color{salmon}{h_{40}}$, $\color{cyan}{h_{60}}$, $\color{yellow}{h_{100}}$, and $\color{pink}{h_{1000}}$ (the yellow is $h_{100}$; it's barely readable on my monitor).

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