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 $B\subset \mathbb{R}^n $.
Assumptions about the set $B$: 1) Assume that $1>P(Z\in B)>0$; 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.
 A: This isn't a full answer, but it does give some more info: I believe the derivative you seek can be negative infinity.  For instance, take the example of $n = 1$ and $B = [-1,1]$.
For legibility, I'll write $1\{A\}$ for the indicator of an event $A$.  Then \begin{align*}
E&\left[Z^2 1\{(X,X+\sqrt{t}Z) \in [-1,1]^2\} \right] \\&= E\left[Z^2 1\{X \in [-1,1]\} 1\left\{Z \in \left[\frac{-1 - X}{\sqrt{t}},\frac{1 - X}{\sqrt{t}} \right] \right\}\right] \\
&= E[Z^2 1_{[-1,1]}(X)] - E\left[Z^2 1\{X \in [-1,1]\} 1\left\{Z \notin \left[\frac{-1 - X}{\sqrt{t}},\frac{1 - X}{\sqrt{t}} \right] \right\}\right]\,.
\end{align*}
I claim that this second term in absolute value is $\Omega(\sqrt{t})$ as $t \to 0$.  To see this, we may bound it below in absolute value by \begin{align*}
E[Z^2 1\{ X \in [1 - \sqrt{t},1] 1\{Z > 0\} ] &\sim \sqrt{t}\cdot\phi(1) E[Z^2 1\{Z > 0\}] \\
&= \sqrt{t} \cdot \phi(1) / 2  
\end{align*}
where $\phi(1)$ is the standard normal density evaluated at $1$.  I suspect an upper bound (in this case) of $\sqrt{t}$ is possible to achieve as well.
EDIT: Some more details on the LB:
\begin{align*} 
E&\left[Z^2 1\{X \in [-1,1]\} 1\left\{Z \notin \left[\frac{-1 - X}{\sqrt{t}},\frac{1 - X}{\sqrt{t}} \right] \right\}\right] \\
&\geq E\left[Z^2 1\{X \in [1-\sqrt{t},1]\} 1\left\{Z \notin \left[\frac{-1 - X}{\sqrt{t}},\frac{1 - X}{\sqrt{t}} \right] \right\}\right] \\
&\geq E\left[Z^2 1\{X \in [1-\sqrt{t},1]\} 1\left\{Z > 0 \right\}\right]
\end{align*}
