Let $f:\mathbb R\to\mathbb C$ be a function that "isn't too discontinuous" and "doesn't grow too rapidly" at $\pm\infty$. Suppose we define the following two "regularized" Fourier transforms of $f$: \begin{align} F_\ell(k) &= \int_{-\ell}^\ell f(x)\, e^{-ikx}\, \frac{dx}{\sqrt{2\pi}} , \qquad \ell>0\\ G_\epsilon(k) &= \int_{-\infty}^\infty f(x)e^{-\epsilon |x|} \,e^{-ikx} \frac{dx}{\sqrt{2\pi}}, \qquad \epsilon>0. \end{align} Each of $F_\ell$ and $G_\epsilon$ is a one-parameter family of functions concocted to make the integrals converge even in $f$ isn't super well-behaved. The first accomplishes this by hard cutoffs on the integrals, and the second does so by taming $f$ through multiplication by a rapidly decaying exponential. When $\ell\to\infty$ and $\epsilon\to 0$, it seems that morally speaking one should obtain the Fourier transform of $f$.
More precisely, I want to say that if $f$ is suitably nice (perhaps something like piecewise smooth and of at most polynomial growth at infinity), then these families are equivalent "regularizations" of the Fourier transform $\hat f$ of $f$ in the sense that for any smooth, rapidly-decaying test function $\varphi$ (i.e. $\varphi$ in Schwartz space $S(\mathbb R)$), we have \begin{align} \lim_{\ell\to\infty}\int_{-\infty}^\infty F_\ell(k)\,\varphi(k)\, dk = \lim_{\epsilon\to 0}\int_{-\infty}^\infty G_\ell(k)\,\varphi(k)\, dk = \int_{-\infty}^\infty \hat f(k)\, \varphi(k)\, dk \end{align} Is this true?
This question is motivated by thinking of $F_\ell$ and $G_\ell$ as families of distributions in which case the equivalence condition above says that these families converge to the same desired tempered distribution -- the Fourier transform of $f$.