Is $\frac{\sin x}{x}$ uniformly continuous on real line $\frac{\sin x}{x}$ is uniformly continuous on a closed interval as it is continuous, but how to extend it to whole real line.
Choosing  what $\delta$ can I proceed?
 A: Hint:
Fix $\varepsilon >0$.
Show that $\frac{\sin x}{x}$ is uniformly continuous on each segment: $\left( - \infty,-M\right) , \left[-M,M\right], \left[M,\infty\right)$ separately. $M$ is so big, that $\frac{\sin x}{x} \approx 0$ on $\left(M,\infty\right), \left(-\infty, -M\right)$ 
A: Set $f(x) = \frac{sinx}{x}$. For a given $\epsilon > 0$, we could find $M >0$, such that $\frac{2}{M} < \epsilon$.
In the closed interval $[-M-1, M+1]$, there exists $\delta >0$ such that $\forall x,y \in [-M-1, M+1]$ and $|x-y| \leq \delta$, we have $|f(x)-f(y)| < \epsilon$.
then take $\delta' = \min\{\delta, 1\}$, we could see that when $|x-y| < \delta'$, we have only three possibilities:


*

*$x,y \in [-M-1, M+1]$

*$x > M, y >M$

*$x < -M, y< -M$


In each case we can get $|f(x)-f(y)|<\epsilon$. For example when $x > M, y >M$, $|f(x) - f(y)| \leq |f(x)|+ f|y| \leq \frac{2}{M} \leq \epsilon$
A: On $[-1,1]$ we have no problem as you say,  
say $x\in [-1,1]^c,\text { now }  |f'(x)|=|{\cos x\over x}+(-{\sin x\over x^2})|\le |{\cos x\over x}|+|{\sin x\over x^2}|\le {1\over |x|}+{1\over x^2}\le 2$
so derivative is bounded so uniformly continuous there too.
