Is $f(x)=x+\frac{x}{x+1}$ uniformly continuous on $(0,\infty)$ 
Is $f(x)=x+\frac{x}{x+1}$ uniformly continuous on $(0,\infty)$

Going from the epsilon delta definition we get: 
$$\forall x,y>1,\text{WLOG}:x>y \ ,\ \forall\epsilon>0,\exists\delta>0 \ s.t. \ |x-y|<\delta\rightarrow |x+\frac{x}{x+1}-y-\frac{y}{y+1}|<\epsilon$$
But I'm not really sure on how to continue from here.
Alternatively: if $\lim\limits_{x\to \infty}f(x)$ exists on $[0,\infty)$ then the function is uniformly continuous, well, it's easy to see here that $\lim\limits_{x\to \infty}f(x)=\infty$ but is that enough ?
 A: Answer here if you want it, I hope that the spoiler tags work. (If not, could someone tell me how to spoiler multiple lines?)

Let $a \in (0,\infty)$ be given. Let $\epsilon >0$ be given. Select $\delta=\frac{\epsilon}{2}$ Then, for all $x \in (0,\infty)$ with $|x-a|<\delta$:$$|f(x) - f(a)|=\left|x+\frac{x}{x+1} - a -\frac{a}{a+1}\right| \leq |x-a| + \left|\frac{x-a}{(x+1)(a+1)}\right|$$$$=|x-a|+\frac{|x-a|}{|x+1||a+1|}\leq |x-a| + |x-a| = 2|x-a| < 2\delta \leq 2\cdot \frac{\epsilon}{2}=\epsilon$$Hence, $f$ is continuous at $a$. Since $a \in (0,\infty)$ and $\epsilon >0$ are given arbitrarily, we have: for all $ a \in (0,\infty)$ and all $ \epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in (0,\infty)$ with $|x-a|<\delta$, $|f(x) - f(a)|<\epsilon$. Conclusion: $f$ is continuous.

A: Take a smaller bite.  Note that a sum of uniformly continuous functions is uniformly continuous.  We know $x\mapsto x$ is uniformly continuous, so we just deal with $x/(x+1)$.
Also, notice that 
$${x\over x + 1 } = 1 - {1\over x + 1}.$$
Now the job is to show that $x\mapsto 1/(x + 1)$ is uniformly continuous.  Can you do the rest?
A: Hint: Use the mean value theorem. Note that, the derivative of the function is bounded on $(0,\infty)$. See a related problem.
