Dominated convergence theorem: a simple application Let $f\in L^1(\mathbb{R})$. Consider the following 
$$
     \lim_{h\rightarrow0}\int_{-\infty}^{+\infty} f(x)e^{-i\zeta x}\,\frac{e^{-ihx}-1}{h}dx
$$
How can I justify that I can pass the limit into the integral? I guess I can use the Dominated convergence theorem but
$$
 \vert f(x)e^{-i\zeta x}\,\frac{e^{-ihx}-1}{h} \vert \le \vert\frac{2}{h}f(x)\vert
$$
and now I can not say that this is less than $\vert 2f(x)\vert$ (which is integrable) because $h$ can be less than $1$. Also, the dominated convergence theorem that I saw was only for limits with $h\rightarrow \infty$ and $h$ an integer.
 A: 
I can not say that this is less than $|2f(x)|$ (which is integrable) because $h$ can be less than $1$.

Indeed. $\frac1h$ is not bounded as $h\to0$, but $\frac{e^{-ihx}-1}{h}$ is! (for fixed $x$)
More specifically, $|e^{z}-1|\leq (e-1)|z|$ for $|z|\leq 1$:
$$\left|\frac{e^z-1}z\right|=\left|\sum_{n=1}^\infty\frac{z^{n-1}}{n!}\right|\leq e-1$$
So $\left|\frac{e^{-ihx}-1}{h}\right|\ll|x|$. Assuming $\int|f(x)x|dx<\infty$ (e.g. if $f$ is a Schwarz function) the Dominated Convergence Theorem applies.

Also, the dominated convergence theorem that I saw was only for limits with $h\to\infty$ and $h$ an integer.

Recall that if $g$ is a function defined in a neighborhood of $0$, then $\lim_{h\to 0}g(h)=L$ iff $\lim_{h_n\to 0}g(h_n)=L$ for all sequences $h_n \to0$ (with $h_n\neq0$). Now let
$$g(h)=\int_{-\infty}^{+\infty} f(x)e^{-i\zeta x}\,\frac{e^{-ihx}-1}{h}dx$$
This idea allows to reduce the limit-integral exchanging issue to the case of sequences (here: $h_n$).
A: $$
\left|
\frac{e^{ihx} - 1}{h} 
\right|
=
\left|
\frac{\left[1 + hx \sum\limits_{k=0}^{\infty} \frac{(h x)^k}{(k+1)!} i^{k+1} \right] - 1}{ h }
\right| 
=
\left|
x \sum\limits_{k=0}^{\infty} \frac{(hx)^k}{(k+1)!} i^{k+1}
\right| 
\leq |x| \sum\limits_{k=0}^{\infty} \frac{|hx|^k}{k!} = |x|e^{|hx|}
$$
