Given $T_n$ is a continuous random variable with the pdf $\ \ f_{T_n}(t) = 8n(1-nt)\ I_{(\frac{1}{2n},\frac{1}{n})}(t)$ for $n\in N$. Find the pdf of the limiting distribution for the sequence $T_1, T_2, \ldots,$
My thought: First, we see that $\lim_{n\rightarrow \infty} f_{T_n}(t) = 0$ for all $t\in R$. Now, we need to compute $P(T_n\leq t)$, but I could not see how to compute this from the given $\ f_{T_n}(t)$. I am not sure if this is correct, but I think $P(T_n\leq t) = \int_{\frac{1}{2n}}^{t} 8n(1-nx)dx = -3+ 8tn - 4n^2t^2$. Thus, $F_{T_n}(t) = -3+ 8tn - 4n^2t^2$ for $t\in (\frac{1}{2n}, \frac{1}{n})$, and $= 0$ otherwise. Now, when taking the limit of $F_{T_n}(t)$ as $n\rightarrow \infty$, we have: $-3 + \lim_{n\rightarrow \infty} 8tn - 4n^2t^2$ for $t\rightarrow 0$, but I don't know how to evaluate this limit.
Could someone please help me with this last step?
Given $f_{T_n}(t) = 8n(1-nt)\ I_{\left(\frac{1}{2n},\frac{1}{n}\right)}(t)$ is positive only for $t\in\left(\frac{1}{2n},\frac{1}{n}\right)$ find that CDF of $T_n$ looks as follows: $$ F_{T_n}(t)=P(T_n\leq t)=\int_{-\infty}^t f_{T_n}(x)\,dx=\begin{cases}0, & t<\frac1{2n}\cr \ldots, & t\in\left(\frac{1}{2n},\frac{1}{n}\right)\cr 1, & t \geq \frac{1}{n}\end{cases} $$
$F_{T_n}(t)=0\to 0$ for all $t<0$. Since $\frac1n \downarrow 0$ as $n\to\infty$, for any strictly positive $t$ there exists $n_0$ such that $F_{T_n}(t)=1$ for all $n\geq n_0$.
We see that for all $t\neq 0$ $$ F_{T_n}(t) \to F_T(t)=\begin{cases}0, & t<0\cr 1, & t\geq 0\end{cases} $$ By definition, it means convergence in distribution to $T=0$ a.s. Limiting distribution is degenerate at $0$. We are not interested in the behaviour of CDF's at the point $t=0$ since this is the discontinuity point of limiting CDF.