Define $f(x) = \begin{cases} \frac{\sin x}{x} & \text{if } x \neq 0\\ 1 & \text{if } x = 0\\ \end{cases}$ Show that $f$ is uniformly continuous (UC) on $\mathbb{R}$.
My Approach: If we can show that $f$ is UC on $[1,\infty]$, then $f$ is UC on $[0,\infty]$. But, notice that $f$ is continuous on $[0,\infty]$. Notice: $f'(x) = \begin{cases} \frac{\cos (x )x - \sin x}{x^2} &\text{if } x \neq 0\\ 0 & \text{if } x = 0\\ \end{cases}$
Since $f$ is differentiable everywhere, then if we can show $f'$ is bounded on $(1,\infty)$, then we will have our desired result. But im stuck on how to prove $f'$ is bounded on the mentioned interval. Can someone help me? ALso i would like to ask if there is a better way to approach this problem...