The goal is to find the limiting distribution of $W_n = Z_n/n^2$ where $Z_n\sim \chi^{2}(n).$
I know from this answer that $W_n\rightarrow 0$ in distribution. I am trying to get this answer using CDFs.
$F_{W_n}(w)=P(\dfrac{Z_n}{n^2}\leq w)=P(Z_n\leq n^2w)=F_{Z_n}(n^2w)$
hence, differentiating partially with respect to $w$ we get $\lim_{n\rightarrow \infty}f_{W_n}(w)=\lim_{n\rightarrow \infty}2n\cdot f_{Z_n}(n^2w)=\lim_{n\rightarrow \infty}2n\cdot \dfrac{\dfrac{1}{2}\cdot e^{-\frac{1}{2}n^2w}\cdot (\frac{1}{2}\cdot n^2w)^{\frac{n}{2}-1}}{\Gamma(\frac{n}{2})}$
I don't know what to do after this.
I know that $\lim_{n\rightarrow \infty}f_{W_n}(w)$ has to be a probability mass function, but the above method will not give a probability mass function for sure. So what am i doing wrong?