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.

$f'(x) = \frac{x\cos (x) - \sin (x) + x^2}{x^2}$ 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.
