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I need to prove that the following function is uniformly continuous on the interval $[0,\infty)$ : $$f(x)=x + \frac{x}{x+1}$$

I want to prove it by defenition, any help?

Watson
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user401516
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    It is sufficient to show that $1/(x+1)$ is uniformly continuous on $[0,\infty)$, equivalently $1/x$ is uniformly continuous on $[1,\infty)$. – Watson Jan 13 '17 at 13:01
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    By the way, after you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. Thank you! :-) – Watson Jan 13 '17 at 13:02
  • (See also http://math.stackexchange.com/questions/1807704) – Watson Jan 13 '17 at 13:07
  • It is Lipshitz-continuous with a derivative bounded by $2$ in absolute value, since $\frac{1}{(1+x)^2}\leq 1$. – Jack D'Aurizio Jan 13 '17 at 13:35
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    @user401516: 14 questions but no accepted answer? Does this mean that our answers do not pass your standards of quality? – Alex M. Jan 13 '17 at 14:10

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Hint: While using definition in $|f(x) - f(y)|$ < $?$ use triangle inequality. Basic idea is to use triangle inequality and take least common multiple of denominators, so $xy - yx = 0$. It will hold because $\frac{x}{x+1} < x$, for all $x > 0$.

Nemanja Beric
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  • Thank you for your answer, but when you say " u/v < u, for u, v > 0 ", I get $1 / 0.5 < 1$ (I applied your claim for $u=1,v=0.5$). Moreover, you could maybe use LaTeX. – Watson Jan 13 '17 at 13:26
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    You are right. u/v < u, for v > 1 and u > 0. In this case it applies (unless I am missing something), because x/(x+1) - y/(y+1) = (x-y)/(x+1)(y+1), and in this case it seames like it holds. Sorry for not using $$. It's a challenge on a phone. – Nemanja Beric Jan 13 '17 at 13:32