The following is the proof from Stein and Shakarchi's Fourier Analysis that given a function $f \in \mathcal{S}(\mathbb{R})$ the Fourier transform of $-2\pi ixf(x)$ is $\frac{d}{d\xi}\hat{f}(\xi)$.
My question is, in the second paragraph of the proof, how do we find a $h_0$ such that for all $|x|\le N$, and $|h|<h_0$, we have $$\Big|\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\Big| \le \frac{\epsilon}{N}.$$ Clearly, for each $x$ we can find a $h_0$ that satisfies the inequality, but to complete the proof, we need to show that we can find a $h_0$ such that the inequality holds uniformly over $|x|\le N$. However, I don't recall any theorems that guarantee such uniform differentiability. I would greatly appreciate any suggestions or solutions.
Observe that $1 - 2\pi i x h\ $ is the sum of the first two terms of the Taylor series for $e^{-2\pi i x h}$. Therefore, $e^{-2\pi i x h} - 1 + 2\pi i x h\ $ is the sum of all except the first two terms: $$e^{-2\pi i x h} - 1 + 2\pi i x h = \sum_{n=2}^{\infty}\frac{(2\pi i x h)^n}{n!}$$ Taking absolute values of both sides and applying the constraint $|x| \leq N$, we obtain $$ |e^{-2\pi i x h} - 1 + 2\pi i x h| \leq \sum_{n=2}^{\infty}\frac{|2\pi x h|^n}{n!} \leq \sum_{n=2}^{\infty}\frac{|2\pi N h|^n}{n!}$$ Divide both sides by $|h| \neq 0$ to get $$\left|\frac{e^{-2\pi i x h} - 1}{h} + 2\pi i x\right| \leq \sum_{n=2}^{\infty}\frac{|2\pi N|^n}{n!}|h|^{n-1} = \sum_{k=1}^{\infty}\frac{|2\pi N|^{k+1}}{(k+1)!}|h|^k$$ The right hand side is a power series with infinite radius of convergence, so certainly it is continuous at the origin $|h| = 0$, where its value is zero. Therefore, as claimed, there is some $h_0$ such that the right hand side will be bounded by $\epsilon/N$ for all $|h| < h_0$.