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Prove/Disprove $f(x) = x + \frac{x}{{x + 1}}$ is uniformly continuous at $\forall x,y \in [0,\infty )$

This is my trial: $$\forall \varepsilon > 0\exists \delta > 0.\forall x,y \in [0,\infty ):$$ Let $$\left| {x - y} \right| < \delta $$ $$\left| {x + \frac{x}{{x + 1}} - y - \frac{y}{{y + 1}}} \right|... \le \delta + \frac{\delta }{{(x + 1)(y + 1)}}$$

As you can see after some algebra I got an expreesion with denominator involving $x,y$.
I need to "get rid" of $x,y$ at the denominator.

One way doing it is just multiplying by $(x+1)(y+1)$, because then I left with just $2\delta$ which is great. what do you think? Am I doing it right?

AndrePoole
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    You can use $f(x)=x+1+\frac1{x+1}$ and use that a sum of continuous functions is continuous. Now, you only have to show that $\frac 1{1+x}$ is continuous on $[0,\infty)$. – Ragnar Dec 26 '13 at 21:19
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    Well, all you need to observe now is that both $x+1$ and $y+1$ are $\geq 1$, hence ..., and so you can just pick $\delta =\epsilon/*$. – Julien Dec 26 '13 at 21:58
  • An intuitive way of seeing this is that $y=x$ is uniformly continuous and $y=\frac {x}{x+1}$ increases at a rate slower than $y=x$, so yes this function is uniformly continuous on $[0,\infty)$ – ireallydonknow Dec 27 '13 at 02:28

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Yes, you're doing it right. $x$ and $y$ are nonnegative so $(x+1)(y+1)\geq 1$, and

$$ \delta + \frac{\delta}{(x+1)(y+1)} \leq 2\delta. $$

So for any $\epsilon>0$, let $\delta = \epsilon/2$, etc., etc.

This is basically julien's comment.

Stefan Smith
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It would be a better choice to estimate derivative first :

$f(x)=x+\frac{x}{x+1}\Rightarrow f'(x)=1+\frac{1}{(x+1)^2}$

check what values would $f'(x)$ takes on $[0,\infty)$

Suppose $|f'(x)|\leq M$..... You can now use definition of derivative :

$|f'(x)|=\lim_{y\rightarrow x}\frac{|f(x)-f(y)|}{|x-y|}$

So, once you know this fellow is bounded by $M$ (If it actually is)

Then, with slight modifications you would end up with something like :

$|f(x)-f(y)|\leq M |x-y|$ for all $x,y\in [0,\infty)$ which would imply uniform continuity... (Why?)

On a serious note, Boundedness of derivative is only to guess what $\delta$ could be for given $\epsilon$.

It may not be good to go for derivative all the time..

Good luck!