Show that the function $f(x) = \frac{\sin x}{x}, x >0$ is uniformly continuous. Show that the function $f\left(x\right) = \frac{\sin x} {x}, x >0$ is uniformly continuous. 
Here what I have done :
Let $x_1, x_2 \in \mathbb R^+$
Then, 
$$|f\left(x_1\right)-f\left(x_2\right) | = \left|\frac{\sin x_1} {x_1} - \frac{\sin x_2} {x_2}\right|
\implies\frac{|x_2\sin x_1 - x_1\sin x_2|}{x_1 x_2}$$
Then, what to do?? 
 A: We have $|\cos x_1-\cos x_2|\le |x_1-x_2|$ for all $x_1, x_2,$ because the absolute value of the derivative is $\le1$. Then
$$\left|\frac{\sin x_1}{x_1}-\frac{\sin x_2}{x_2}\right|=\left|\int^1_0(\cos tx_1-\cos tx_2)\,dt\right|\le|x_1-x_2|\int^1_0t\,dt=|x_1-x_2|/2.$$  
A: The function extends by continuity to $x=0$ by setting $f(0)=1$. Thus $f$ is uniformly continuous on the compact interval $[0,1]$. It remains to show that $f$ is uniformly continuous on $[1,\infty)$. This can be done for instance by showing that the derivative on this interval is bounded.
A: Here is a way you can do this. 
For $x\in (1,\infty)$
First, note that if $f=\frac{sin(x)}{x}$, then the derivative is $f'(x)=\frac{cos(x)}{x}-\frac{sin(x)}{x^2}$. The derivative $f'$ is actually bounded,  You can use the crude bound $|f'|<1$. If the derivative is bounded, the function $f$ is Lipschitz, which also means it is uniformly continuous. 
for $x\in (0,1)$ there is more work to be done, as the derivative is not bounded, and there are uniformly continuous functions with unbounded derivatives. 
For uniform continuity, we want to show that $\forall \epsilon>0$, we have a $\delta>0$ with $|f(x_1)-f(x_2)|<\epsilon$ when $|x_1-x_2|<\delta$. For $f$ to be uniformly continuous, the choice of $\delta$ cannot depend on $x$. In other words, there is a universal $\delta$ $\forall x$. 
Also, use the fact that if $g,h$ are uniformly continuous, then their sum $f+g$ is uniformly continuous. Define $g(x)=\frac{cos(x)}{x}$ and $h(x)=\frac{-sin(x)}{x^2}$. For $g$ and $h$, choosing $\delta=\frac{\epsilon}{2}$ should do the trick (unless I was sloppy). Now just put $f=g+h$ and conclude the sum of two uniformly continuous functions are uniform to complete the proof. 
