Using (
18), we can write
where
At a high level, the proof will follow a similar overall approach as in the derivation in
Section 3.1. We will first invoke the dominated convergence theorem evaluate the limit of the
term directly, as well as to show that the quantities
and
each approach zero as
We will then use L’Hôpital’s rule and Leibniz’s theorem on differentiation under the integral sign to evaluate the limits of the
and
terms. Let
be a set of Lebesgue measure 1 such that
is continuously differentiable everywhere on
. Then
and
are bounded almost everywhere on
(see, for example, [28]) and therefore, almost everywhere on
Let
and
be these upper bounds on
and
respectively. Then, for
the integrands in (40), (41), and (42) are bounded (uniformly in
) by
and
respectively, each of which is a finite constant and therefore yields a finite integral on
Thus, using the dominated convergence theorem, we have
Similarly, we have
since
and
We then have, using L’Hôpital’s rule,
where
follows from Leibniz’s theorem, since the integrand is continuously differentiable in both
and
u,
follows from the dominated convergence theorem (taking the limit inside the integral), and
follows from the substitution
Following very similar steps, we have
Finally, using (39), we have
which establishes the limit for
Here,
follows by plugging in (43), (44), and (45). The limit of
can be demonstrated through a similar sequence of steps. □