= Solution
Let $X_t$ be the number of telephone lines in use. The <memorylessness of the exponential distribution> makes $(X_t)_{t\geq0}$ a <continuous-time Markov chain>, specifically the <M-M-infinity queue> with generator rates
$$
q_{n,n+1}=\lambda,
\qquad
q_{n,n-1}=n\mu,
\qquad
q_{n,n}=-(\lambda+n\mu).
$$
Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the <exponential survival probability> $p_t$, independently of the others. Their surviving number is therefore
$$
Y_t\sim\operatorname{Binomial}(n,p_t).
$$
A call arriving at time $u\in[0,t]$ survives until $t$ with probability $e^{-\mu(t-u)}$. By <Poisson thinning>, the number of surviving new calls is independent of $Y_t$ and has law
$$
Z_t\sim\operatorname{Poisson}\!\left(
\lambda\int_0^te^{-\mu(t-u)}\,du
\right)
=\operatorname{Poisson}\!\left(
\frac\lambda\mu(1-p_t)
\right).
$$
Hence $X_t=Y_t+Z_t$ in distribution. Multiplying the <probability generating function>[probability generating functions] of the two <independent random variables> gives, for $-1\leq s\leq1$,
$$
\boxed{
\mathbb E[s^{X_t}]
=(1-p_t+p_ts)^n
\exp\!\left\{
\frac\lambda\mu(1-p_t)(s-1)
\right\}.}
$$
Equivalently, the <transient distribution of an M-M-infinity queue> is
$$
\boxed{
X_t\overset{d}{=}
\operatorname{Binomial}(n,e^{-\mu t})
+\operatorname{Poisson}\!\left(
\frac\lambda\mu(1-e^{-\mu t})
\right),}
$$
with independent summands. As $t\to\infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda/\mu$. Therefore
$$
\boxed{X_t\xrightarrow d\operatorname{Poisson}(\lambda/\mu),}
$$
which proves the stated large-time approximation.
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