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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?

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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 $[1,\infty)$. – 2017-01-13
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    (See also http://math.stackexchange.com/questions/1807704) – 2017-01-13
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    It is Lipshitz-continuous with a derivative bounded by $2$ in absolute value, since $\frac{1}{(1+x)^2}\leq 1$. – 2017-01-13
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    @user401516: 14 questions but no accepted answer? Does this mean that our answers do not pass your standards of quality? – 2017-01-13

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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 $x > 0$.

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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. – 2017-01-13
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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. – 2017-01-13