= Solution
Continue with $A_d$ equal to the integral defined in the PDF and $u(z)=|z|^{2-d}$. Until the <Brownian motion> hits the radius-$\varepsilon$ ball, $u(B_s)$ is a bounded <local martingale>, hence a true <martingale>. At the finite <stopping time> $t\wedge T_\varepsilon$, <optional stopping> gives
$$
u(x)=\varepsilon^{2-d}\mathbb P_x(T_\varepsilon\leq t)+\mathbb E_x[u(B_t);T_\varepsilon>t].
$$
Rearranging,
$$
\varepsilon^{2-d}\mathbb P_x(T_\varepsilon\leq t)
=u(x)-\mathbb E_xu(B_t)+\mathbb E_x[u(B_t);T_\varepsilon\leq t].
$$
The fixed random variable $u(B_t)$ is integrable by the <Gaussian heat kernel> and the local-integrability calculation above. Also $\mathbb P_x(T_\varepsilon\leq t)\leq(\varepsilon/|x|)^{d-2}\to0$. For any $R>0$, the final <expectation> is at most
$$
\mathbb E_x[u(B_t);u(B_t)>R]+R\mathbb P_x(T_\varepsilon\leq t).
$$
First send $\varepsilon\downarrow0$, then $R\to\infty$, proving that it vanishes. The identities from part (b) give
$$
u(x)=A_d^{-1}\int_0^\infty p(s,0,x)ds,\qquad
\mathbb E_xu(B_t)=A_d^{-1}\int_t^\infty p(s,0,x)ds.
$$
Subtracting yields the <small-ball Brownian hitting asymptotic>
$$
\boxed{\lim_{\varepsilon\downarrow0}\varepsilon^{2-d}\mathbb P_x(T_\varepsilon\leq t)
=A_d^{-1}\int_0^t p(s,0,x)\,ds.}
$$
Thus the constant in this printed limit requires exactly the same reciprocal correction as in part (b). For $d=3$ the multiplier is $2\pi$, not $1/(2\pi)$. The argument proves the corrected finite-time limit directly, rather than inferring it from the eventual hitting probability alone.
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