Let $\{X_n\}_{n=1}^\infty$ and $X$ be absolutely continuous random variables on a probability space $(\Omega,\mathcal{F},P)$, with density functions $f_{X_n}(x)$ and $f_X(x)$. Are there any hypotheses on $\{X_n\}_{n=1}^\infty$ and/or $X$ that guarantee the convergence $\lim_n f_{X_n}(x)=f_X(x)$ for a.e. $x\in\mathbb{R}$?
The most related result that I know is the following: if $\{X_n\}_{n=1}^\infty$ does not converge in law to $X$, then it not possible to have $\lim_n f_{X_n}(x)=f_X(x)$ for a.e. $x\in\mathbb{R}$, by Scheffé's Lemma. But this result does not give a sufficient condition for $\lim_n f_{X_n}(x)=f_X(x)$.
First see that $F_n$, the distribution function of $X_n$, weakly converges to $F$, the distribution function of $X$, i.e., $F_n\stackrel{w}{\to} F$ if and only if $\lim_{n\to\infty} F_n(x)=F(x)$ for all points of continuity of $F$. (Williams, page 181)
Since $X$ is an absolutely continuous RV then for all $x\in\mathbb R$, $\displaystyle\lim_{n\to\infty} F_n(x)=F(x)$. The question boils down to see under which condition the point-wise convergence implies the convergence of derivatives. In general, the point-wise convergence does not imply the convergence of derivatives.
Note that the convergence of derivatives is equivalent to the exchange of limits in the following: $$ \lim_{n\to\infty}\lim_{y\to x}\frac{F_n(y)-F_n(x)}{y-x}=\lim_{y\to x}\lim_{n\to\infty}\frac{F_n(y)-F_n(x)}{y-x} $$
The sufficient condition for this exchange of limits usually involves some notion of uniform convergence. Some of those conditions are discussed here.
For instance Corollary A of the paper implies that for random variables supported on the interval $[a,b]$, the sequence of densities $f_n(x)$ converges uniformly to $f(x)$ if and only if the sequence $f_n(x)$ is uniform on every point of $[a,b]$.
NOTE: The sequence $f_n$ is said to be uniform at $x$ if for every $\epsilon > 0$, there exists a $\delta> 0$ and an $N$ such that if $|y — x|<\delta$ and $m,n>N$, then $\mid f_m(y) —f_m(x) -f_n(y)+f_n(x)\mid <\epsilon$.