Uniform continuity of $f(x)=x+\frac{\sin x}{x}$ in $(0,\infty)$ 
Prove the following function is uniformly continuous: $f(x)=x+\frac{\sin x}{x}$ in $(0,\infty)$.

I thought of trying to show that the derivative is bounded thus the function is lipschitz and we're done but the derivative isn't bounded. 
Second try: I know that if $f,g$ are continuous then $f+g$ is continuous as well. $x$ is known to be uniformly continuous, so we're left with showing that $\frac{\sin x}{x}$ is lipschitz  which is easy and we're done. 
Is my approach correct ? 
Thanks.
 A: You know $f_1(x) \equiv x$ is uniformly continuous on $(0,\infty)$.   You need to show $f_2(x) \equiv \sin(x)/x$  is uniformly continuous on $(0,\infty)$.
All you need to do is show $f_2$ can be extended continuously to $[0,\infty)$ and  show $f_2(x) \to 0$ as $x \to \infty$.  It follows easily that $f_2$ is uniformly continuous on $[0,\infty)$ (use the fact that a continuous function on a compact interval is uniformly continuous on that interval).  In fact, any continuous function $g:[0,\infty)\to \mathbb{R}$ with a finite limit $\lim_{x\to \infty}f(x)$ is uniformly continuous on $[0,\infty)$.
HINT: let $\epsilon > 0$.  Let $a > 0$ be big enough so that for $x_1, x_2 \geq a$, $|f_2(x_1)-f_2(x_2)| < \epsilon/2$.  $f_2$ is uniformly continuous on the compact interval $[0, a+1]$.  Now you have to find $\delta > 0$ so that if $x_1, x_2 \geq 0$ with $|x_1-x_2| < \delta$, then $|f_2(x_1)-f_2(x_2)|<\epsilon$. HINT: if $\delta< 1$, then $x_1$ and $x_2$ are both in $[0,a+1]$ or both in $[a,\infty)$. 
So you don't need to find, use, or estimate the derivative of $\sin(x)/x$. That derivative actually is bounded, but you don't need that property (by the way, there are uniformly continuous functions on $[0,\infty)$ with unbounded derivative, such as $\sin(e^x)/(x^2+1)$).
