Suppose a function $f$ is of rapid decrease. Then, we have $\displaystyle -2\pi ixf(x)\longrightarrow \frac{d}{d\xi}\hat f(\xi)$
It suffices to show $\displaystyle\int_{-\infty}^{\infty}\left\vert f(x)\right\vert\left\vert\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\right\vert dx\to0$ as $h\to0$
By definition of rapid decrease, we have: for all $\epsilon>0$, there exists $N\in\mathbb{N}$ s.t $$\displaystyle\int_{|x|\geq N}\left\vert x\right\vert\,dx\leq \epsilon\textrm{ and }\int_{|x|\geq N}|x|\left\vert f(x)\right\vert\,dx\leq\epsilon$$ Moreover, there exists $h_{0}$ s.t whenever $|h|<h_{0}$, we have: $$\left\vert\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\right\vert\leq\frac{\epsilon}{N}\textrm{ $\,\,\,\,\,\,\,\,\,\,$I am so confused by this step!}$$
Can somebody tell me the detail behind the inequality above?
We are concerned with
$$\left\vert\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\right\vert\leq\frac{\epsilon}{N}.$$
The short answer is that this is the derivative of $e^{-2\pi i x y}$ in terms of $y$, at $y = 0$. Let's expand on that a bit.
Call $g(y) := e^{-2\pi i x y}$, where I'm thinking of $x$ as fixed. Then this is just the ordinary exponential function, and is smooth and, more to the point, differentiable. We know that $g'(y) = (-2\pi i x)e^{-2\pi i x y}$. What does this mean? This means that $$ \lim_{h \to 0} \frac{g(h) - g(0)}{h} = g'(0),$$ or equivalently, $$ \lim_{h \to 0} \frac{e^{-2\pi i x h} - 1}{h} + 2 \pi i x = 0.$$ The existence of $h_0$ for each $\epsilon, N$ is a restatement of the definition of the limit.