The pdf of last interval of a Poisson process in $[0,T]$ 
Assume a Poisson point process with rate $\lambda$ in time $[0,T]$. Supoose $X$ is the random variable representing the time between the last arrival and $T$. What is the probability density function of $X$ as $T\to \infty$?

The pdf is $\frac{d}{dx}P\left(X\leq x\right)$. We can break up $P\left(X\leq x\right)$ by the number of arrivals in time $[0,T]$:
\begin{align}
P\left(X\leq x\right)&=\sum_{k=1}^{\infty}Poisson(k, \lambda T)P\left(X\leq x|k   \text{ arrivals}\right)\\
&=\sum_{k=1}^{\infty}Poisson(k, \lambda T)P\left(T-T_k\leq x|k   \text{ arrivals}\right)\\
&=\sum_{k=1}^{\infty}Poisson(k, \lambda T)P\left(T-T_k\leq x|T_k \leq T <T_{k+1}\right)
\end{align}
where $T_k=\sum_{i=1}^{k}A_i$ is the time $k^{th}$ arrival, and $A_i$ is the $i^{th}$ inter-arrival time. Any idea how to continue? Or, a resource that already has the answer?
 A: I'll denote $T_k = A_1 +\dots + A_k$.
If there are exactly $k$ arrivals in time $T$, that means $T_k\leq T < T_{k+1}$.
So you can write  \begin{align}P(T_k\geq T-x\mid \text{k arrivals})&=\frac{P(T_k\geq T-x, \ T_k\leq T < T_{k+1})}{P(\text{k arrivals})}\\&=\frac{P(T-x \leq T_k\leq T < T_{k+1})}{P(\text{k arrivals})}\\&=\frac{P(T-x \leq T_k\leq T < T_k+A_{k+1})}{P(\text{k arrivals})}\end{align}
From here, you can use the independence of $A_{k+1}$ and $T_k$ and try to work out the numerator as the integral $$\iint\limits_{\{T-x \leq u\leq T < u+v\}}f_{T_k}(u)f_{A_{k+1}}(v)\,du\,dv,$$noting that $T_k\sim Gamma(k,\lambda)$ and $A_{k+1}\sim Exp(\lambda)$.

On the other hand, there is an easier way to approach the problem.
Denote $N(t)$ the number arrivals in time $t$. Then $T_{N(T)}$ is the time of the last arrival in the interval $[0,T]$. So, for $x\in (0,T)$, we have:
\begin{align}P(X\leq x)&=P(0\leq T-T_{N(T)}\leq x)\\&=P(T-x\leq T_{N(T)}\leq T)\\&=P(N(T)-N(T-x)\geq 1)\\&=1-P(N(x)=0)\\&=1-e^{-\lambda x},\end{align}
where the last two equations follow from $N(T)-N(T-x)$ having the same distribution as $N(x)$, which is $Poisson(\lambda x)$.
Thus we get the cdf of $X$ (note that $X$ is a.s. positive and $X\leq T$ a.s.):
$$F(x)=\begin{cases}0,&x\leq0\\
1-e^{-\lambda x},& x\in(0,T)\\
1,& x\geq T
\end{cases}$$
This looks like the cdf of the exponential distribution, but we have a jump at $x=T$, so it's not really the exponential distribution. But when $T\to \infty$ as the question said, we get the exponential distribution.
