= Solution
The <size-biased distribution> associated with $\xi_1$ is defined by
$$
\mathbb E h(\widehat\xi_1)
=\frac{\mathbb E[\xi_1h(\xi_1)]}{\mathbb E\xi_1}
=\lambda\mathbb E[\xi_1h(\xi_1)]
$$
for every bounded measurable $h$. Thus, if $\xi_1$ has density $f$, then $\widehat\xi_1$ has density $\lambda x f(x)$.
Now suppose that $\xi_1$ is exponential with rate $\lambda$. The renewal epochs form a <Poisson process>. Write
$$
A_t=t-S_{N_t},\qquad R_t=S_{N_t+1}-t,
$$
so that $L(t)=A_t+R_t$. Independent stationary increments and the exponential waiting-time law show that $R_t$ is exponential with rate $\lambda$, independent of the history up to $t$ and hence of $A_t$. For fixed $a\geq0$ and $t>a$,
$$
\mathbb P(A_t>a)
=\mathbb P(N_t-N_{t-a}=0)
=e^{-\lambda a}.
$$
Consequently $A_t$ converges in distribution to another rate-$\lambda$ exponential variable, independent of $R_t$. Therefore $L(t)$ converges to the sum of two independent rate-$\lambda$ exponentials, whose density is
$$
\int_0^x\lambda e^{-\lambda y}\lambda e^{-\lambda(x-y)}\,dy
=\lambda^2xe^{-\lambda x}.
$$
The size-biased density of $\xi_1$ is likewise
$$
\lambda x\big(\lambda e^{-\lambda x}\big)
=\lambda^2xe^{-\lambda x}.
$$
Hence the <exponential renewal interval limit> is
$$
\boxed{L(t)\xrightarrow{d}\widehat\xi_1}.
$$
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