How fast does $lim_{ t to 0} E left[ |Z|^2 1_{B}(X,X+sqrt{t} Z) right]= E left[ |Z|^2 right] E[1_B(X)]$

Let $X in mathbb{R}^n $ and $Z in mathbb{R}^n $ be two independent standard normal random vectors.

We are interested in the following quantity:
begin{align}
E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right]
end{align}
for some set $Bsubset mathbb{R}^n $.

Assumptions about the set $B$: 1) Assume that $1>P(Zin B)>0$; 2) (Optional) $B$ is convex.

Concretely, we are interested in how this quantity behaves as $t to 0$.

First, it is easy to show that
begin{align}
lim_{ t to 0} E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right]= E left[ |Z|^2 right] E[1_B(X)],
end{align}
where we have used the dominated convergence theorem and the bound $|Z|^2 1_{B times B}(X,X+sqrt{t} Z) le |Z|^2$.

My question is: Can we say something about how fast does this approach the limit? Specificaly, can we say something about
$$lim_{ t to 0} frac{d}{dt} E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right]= ???$$

Edit: The derivative is given by
begin{align}
&2 frac{d}{dt} E left[ |Z|^2 1_{B times B}(X,X+sqrt{t} Z) right]\
&=frac{E[|Z|^4 1_{B times B}(X,X+sqrt{t} Z)]- (n+2) E[|Z|^2 1_{B times B}(X+sqrt{t} Z ,X) ]}{t}
end{align}

Now, if take the limit as $t to 0$ of the numerator than we get
begin{align}
&lim_{n to infty} E[|Z|^4 1_{B times B}(X,X+sqrt{t} Z)]- (n+2) E[|Z|^2 1_{B times B}(X+sqrt{t} Z ,X) ]\
&= E left[ |Z|^4 right] E[1_B(X)] – (n+2) E left[ |Z|^2 right] E[1_B(X)]\
&=0
end{align}
where we have used that the fourth moment is given by $E left[ |Z|^4 right]=n(n+2)$.

Therefore, we have zero over zero.
I tried using L’hospital rule more times, but we keep getting zero over zero no matter how many times we apply L’hospital rule.