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Proof that the function $f: [0, \infty) \ni x \mapsto \frac{x^{2}}{x + 1} \in \mathbb{R}$ is uniformly continuous.

On the internet I found out that a function is uniformly continuous when

$\forall \varepsilon > 0 \ \exists \delta > 0: \left | f(x)-f(x_{0}) \right | < \varepsilon$ whenever $\left | x - x_{0} \right | < \delta .$

Because I don't know how to prove it calculative, I have drawn the function and showed its uniform continuity like that. But I'd like to know how to do it the other, more professional and efficient way. I've watched some videos but anyway couldn't find a solution. Also tried to for almost 2 hours myself but nothing came out, too.

For the drawing, I think there is actually a mistake, at the beginning the epsilon (first one I haven't drawed) seems smaller than the others, while the others have same size. (In addition I skipped the other part of the function because it's trivial).

Here is the picture:

enter image description here

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The best way to start these types of problems is to start by messing with the part $|f(x) - f(y)| < \epsilon$ of the definition. Note that, by combining fractions and multiplying everything out we have

$f(x) - f(y) = \frac{x^2}{x+1} - \frac{y^2}{y+1} = \frac{x^2y-y^2x+x^2-y^2}{xy+x+y+1}$.

After playing around with some grouping I found that this can be rewritten as

$\frac{xy(x-y)+(x^2-y^2)}{xy+x+y+1} = \frac{(x-y)(xy+x+y)}{xy+x+y+1}$.

As a hint for where to go from here it is important to remember that $x, y \in [0, \infty)$.

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  • $\begingroup$ Thank you very much, this should be a good start to find an end for me. Im really happy on maths section there aren't as many a-holes as in programming section. Thank you all! $\endgroup$ – roblind Jun 18 '16 at 15:48
  • $\begingroup$ I got the solution, if anyone wants see it I can post it here. It will be a pain for me to type it (new to Latex) but if it's necessary, let me know :P $\endgroup$ – roblind Jun 18 '16 at 16:25
  • $\begingroup$ Your welcome. Yeah I think people into math are generally nicer or at least that's been my experience. Its probably because we always learn something new or get better at it by helping people out with math. You can post it if you'd like us to check it. $\endgroup$ – Jeff L. Jun 19 '16 at 2:15
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Hint:

$f(x)=\frac{x^2}{x+1}$ is the sum between two uniformly continuous functions over $\mathbb{R}^+$: $x-1$ and $\frac{1}{x+1}$.

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  • $\begingroup$ I think problem is I also need proof that both these functions you said are uniformly continuous? I think not allowed to just say they are. But not sure.. $\endgroup$ – roblind Jun 18 '16 at 13:13
  • $\begingroup$ @Hamudii: further hint: both are Lipschitz-continuous functions. $\endgroup$ – Jack D'Aurizio Jun 18 '16 at 13:20
  • $\begingroup$ Ok then I think I cannot use your way because we haven't discussed it (we may not use things we haven't discussed; never heard of Lipschitz, just Leibniz ^^) :( $\endgroup$ – roblind Jun 18 '16 at 13:36
  • $\begingroup$ @Hamudii: a differentiable function with a bounded derivative is uniformly continuous. Call it as you like, but it is a trivial fact. $\endgroup$ – Jack D'Aurizio Jun 18 '16 at 13:41

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