Prove uniform contininuity (probably by Lipschitz continuity) Prove uniform continuity at $(0,\infty)$ for:
$$f(x) = x + \frac{\sin (x)}{x}$$
Derivative is:
$$f'(x) = \frac{x\cos (x) - \sin (x) + x^2}{x^2}$$
so, taking the limit at $\infty$ I got the value of $1$.
Looking at the graph I see infinite numbers of maximum points converges to $1$ of course.  
Wolfram-alpha is telling me there is no global maximum (this is kinda weird, isn't it?).
Anyway, what should I do next? by looking at the graph it is sure can be bounded by a line. I just don't know how to describe it mathematically.
 A: $f'(x) = \frac{x\cos (x) - \sin (x) + x^2}{x^2}$ can be extended to a continuous function on $\mathbb{R}$ (check that $\lim _{x \rightarrow 0}f'(x)=1$). Thus it is bounded on the interval $[-1,1]$.
To show that $f'$ is bounded on $\mathbb{R}$ it remains to show that it is bounded on $[1,\infty]$ and $[-\infty,-1]$.
Let us first consider $[1,\infty]$, for $x\geq1$ we have that 
$$\frac{x\cos (x) - \sin (x) + x^2}{x^2}\leq\frac{x^2+x}{x^2}\leq\frac{2x^2}{x^2}=2$$
It is equally easy to see that $f'$ is bounded on $[-\infty,-1]$. I leave this case for you to work out.
A: TZakrevskiy provided the critical observation in his comment, but I'll try to expand on why it's a natural perspective to take here. Intuitively, your function $f(x)$ is well-behaved near $0$ and near $\infty$ but in different ways. Near the origin, you're essentially looking at the function $\sin x /x$, which you probably are aware can be extended to a continuous function on $[0,\varepsilon]$ for any $\varepsilon$.
At the same time, you've mentioned that $f'(x)$ is bounded as $x \to \infty$ and hence uniformly continuous on $[\varepsilon,\infty)$ for sufficiently large $\varepsilon$ (and in fact for any $\varepsilon > 0$ since $f'$ is continuous$).
Accordingly, a proof of the uniform continuity of $f$ should be broken into two parts, where you establish the uniform continuity of $f$ near $0$ and $\infty$ separately. One way to make this precise is sketched as follows:


*

*Note $f$ extends to a continuous function $\hat f$ on $[0,\infty)$, which I'll now refer to as $f$ from now on.

*Pick you favorite $\varepsilon \in (0,\infty)$ (any will do). You know without any work that $f$ is uniformly continuous on $[0,\varepsilon]$ (why?). (BTW: This is why we extend $f$.)

*Show that $f'$ is bounded on $[\varepsilon,\infty)$, and conclude accordingly that $f$ is uniformly continuous on $[\varepsilon,\infty)$. (BTW: If you haven't seen this proof, try to proof bounded derivative $\implies$ u.c. on your own. It's a great exercise!)

*Conclude that $f$ is continuous on $(0,\infty) \subset [0,\varepsilon]\cup[\varepsilon,\infty)$. (If you haven't explicitly thought about why this is, do so now.)

