# How can average queuing delay exceed 1 if the $La/R \leq 1$?

Consider a router with a packet arrival rate of $a$ where each packet has $L$ bits, and a transmission rate (in bits/s) of $R$. My textbook describes the traffic intensity as $\frac{La}{R}$ and states that as $\frac{La}{R} \to 1$, the average queueing delay approaches $\infty$. I'm just struggling to understand why this is the case. Is it because this increases the chance that $La$ exceeds $R$ during periods of time, and that long queues can accumulate?

For example, suppose the arrival and service process are all deterministic process. For example, a packet of one bit ($L=1$) arrives in every second ($a=1$), and the router transmits a bit in every second ($R=1$). Of course that in such case the average queuing delay equals to 1.
However, if for example in a system where the packet arrivals are determined by a Poisson process with parameter of $\lambda=a$ seconds and job service times have an exponential distribution (M/M/1) with average of $1/\mu=\frac{L}{R}$ seconds (which describes a system with packet length $L$ and average transmission rate of $R$), i.e., $\mu=\frac{R}{L}$ the service time equals to $\frac{1}{\mu-\lambda}- \frac{1}{\mu}$. As $\frac{La}{R}$ approaches $1$, then $\mu-\lambda$ approaches $\infty$ the service time approaches infinity.
Now for the strange behavior of $\frac{La}{R}=1$ for M/M/1 queue: When $\frac{La}{R}=1$ that means that the probability a packet enters the router equals the probability that a packet exits the router. That means, the probability that the router contains $i$ packets equals the probability that the router contains $j$ packets, for every $i$ and $j$. Thus, the number of packets in router approaches $\infty$, as well as the queuing delay.