Obs.: I'm studying Fourier analysis but this is more a real-analysis/limits/suprema-infima question.
Let $\mathcal S$ be the Schwartz space on $\mathbb R$, i.e., the set of functions $f\in\cal C^{\infty}$ on $\mathbb R$ such that for every $k,l\ge 0$ $$ \sup |x|^k|f^{(l)}(x)|<\infty. $$ The Fourier transform of a function $f\in\mathcal S$ is defined by $$ \hat f(\xi) := \int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}\space dx. $$
Proposition: The Fourier transform of $f'$ is the function $\xi\mapsto 2\pi i\xi\hat f(\xi)$.
The proof I'm reading goes by the following:
$$ \int_{-N}^Nf'(x)e^{-2\pi ix\xi}dx = \left[f(x)e^{-2\pi ix\xi}\right]_{-N}^N - \int_{-N}^Nf(x)e^{-2\pi ix\xi}(-2\pi i\xi) dx = f(N)e^{-2\pi iN\xi}-f(-N)e^{-2\pi i(-N)\xi} + 2\pi i\xi\int_{-N}^Nf(x)e^{-2\pi ix\xi}dx $$
Then by making $N\to\infty$ we get the result.
Here I just don't know why $f(N)e^{-2\pi iN\xi}-f(-N)e^{-2\pi i(-N)\xi}$ goes to $0$. I only know how to use that $f\in\cal S$ to get that this expression is bounded, but I don't see how it guarantees that it goes to zero.
Thanks in advance.