# simple calculus inquiry

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...

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I would rather proceed from the first principles. Here is the sketch. Consider first $x,x_0\in(1,\infty)$ $$\left|\frac{\sin x}{x}-\frac{\sin x_0}{x_0}\right|=\left|\frac{x_0\sin x-x\sin x_0}{xx_0}\right|=\left|\frac{x_0\sin x -x\sin x+x\sin x-x\sin x_0}{xx_0}\right|\le\\\left|\frac{1}{x_0}\left(\sin x-\sin x_0\right)\right|+\left|\frac{1}{x_0}\frac{\sin x}{x}\left(x- x_0\right)\right|<\left|\sin x-\sin x_0\right|+\left|x-x_0\right|=\\\left|2\sin\frac{x-x_0}{2}\cos\frac{x+x_0}{2}\right|+\left|x-x_0\right|\le\left|2\sin\frac{x-x_0}{2}\right|+\left|x-x_0\right|\le 2\left|x-x_0\right|$$ Hence given $\epsilon>0$, take $\left|x-x_0\right|<\delta=\frac{\epsilon}{2}$.
Now on $(0,1)$ a similar argument should be possible after letting $x=\frac{1}{t}$; and $x=1$ can be dealt with separately.