Given $f$ a Lebesgue integrable function on $\mathbb{R}$ with finite $L^1$-norm, I am asked to show that $\lim_{n\to\infty} \|f_n - f\|_1 = 0$, where $f_n = f \ast g_n$ and $g_n = \sqrt{\frac{n}{2\pi}}e^{-nx^2/2}$.
A first thought was to use that $\int_\mathbb{R}\int_\mathbb{R} f\ast g = (\int_\mathbb{R} f)(\int_\mathbb{R} g)$, but am unable to manipulate the integrand into a form where that property of convolution is useful:
$$\begin{align} \|f_n - f\|_1 &= \int_\mathbb{R} |f_n(x) - f(x)| \, dx \\ &= \int_\mathbb{R} \bigg| \left( \int_\mathbb{R} g_n(x - y) f(y) \, dy \right) - f(x) \bigg| \, dx \end{align}$$
I do know that for each $n$, $\|g_n\|_1 = 1$ (it is a normalized Gaussian integral). On the other hand, $\lim_{n\to\infty} g_n(x) = 0$ for all $x \in \mathbb{R} \setminus \{0\}$; since both $f$ and $g_n$ are integrable, the inner integral on the last line above would go to zero by the Lebesgue dominated convergence theorem, wouldn't it? But that would give $\lim\|f-f_n\|_1 = \|f\|_1$.
It feels as though I am approaching this problem from the wrong direction and am unsure of how to proceed.