Find $ \lim_{ \epsilon \to 0} \frac{1}{R^n-(R-\epsilon)^n} \int_{R-\epsilon\le \|x\| \le R} e^{-\frac{\|x-\mu\|^2}{2}} dx$ How to find the following limit
\begin{align}
 \lim_{ \epsilon \to 0} \frac{1}{R^n-(R-\epsilon)^n} \int_{R-\epsilon\le \|x\| \le R} e^{-\frac{\|x-\mu\|^2}{2}} dx.
\end{align}
For the case of $n=1, $ I was able to compute this and the limit is given by  $e^{-\frac{(\mu -R)^2}{2}}+e^{-\frac{(\mu +R)^2}{2}}$. However, not sure how to do it in general. 
Reasoning that I had (which is wrong)
Intuitively for a very small epsilon 
\begin{align}
\int_{R-\epsilon\le \|x\| \le R} e^{-\frac{\|x-\mu\|^2}{2}} dx&= e^{-\frac{\|R-\mu\|^2}{2}}  \int_{R-\epsilon\le \|x\| \le R}  dx\\
&= e^{-\frac{\|R-\mu\|^2}{2}}  V_n  (R^n-(R-\epsilon)^n),
\end{align}
where $V_n$ is the volume of a unit ball.  So, my guess for the limit was
\begin{align}
e^{-\frac{\|R-\mu\|^2}{2}}  V_n .
\end{align}
However, this does not appear to be correct as it does not agree with the $n=1$ case.  As Did pointed out my assumption is not correct. 
This question looks possibly related to 
\begin{align}
\lim_{\epsilon \to 0} \frac{1}{Vol(B_c(\epsilon))} \int_{B_c(\epsilon)} f(x) dx ,
\end{align}
where  $B_c(\epsilon)$ is a ball of radius $\epsilon$ centered at $c$. The difference is that we are looking at the annulus instead of a ball. 
 A: First, let's evaluate the limit. Note that it's an indeterminate form
$$\lim_{\epsilon\to 0}\frac{p(\epsilon)}{q(\epsilon)}$$
where
$$\begin{split}
q(\epsilon)&\stackrel{\text{def}}{=}R^n-(R-\epsilon)^n\\
&=R^n-(R^n-nR^{n-1}\epsilon +O(\epsilon^2))\\
&=nR^{n-1}\epsilon +O(\epsilon^2)
\end{split}$$
$$\begin{split}p(\epsilon)&\stackrel{\text{def}}{=}\int_{R-\epsilon\le \|x\| \le R} \mathrm{e}^{-\frac{\|x-\mu\|^2}{2}} dx\\
&=\int_{R-\epsilon}^R\int_{S_{n-1}} \mathrm{e}^{-\frac{\|r\hat{x}_{\Omega}-\mu\|^2}{2}}r^{n-1}\mathrm{d}\Omega\,\mathrm{d}r\text{.}\end{split}$$
($\hat{x}_{\Omega}$ is the direction vector of $x$, and $\Omega$ ranges over the unit sphere $S_{n-1}$.) Let $$f(r)\stackrel{\text{def}}{=}\int_{S_{n-1}} \mathrm{e}^{-\frac{\|r\hat{x}_{\Omega}-\mu\|^2}{2}}r^{n-1}\mathrm{d}\Omega\text{.}$$ Then
$$\begin{split}p(\epsilon)&=\int_{R-\epsilon}^Rf(r)\,\mathrm{d}r\\
&=\int_{0}^{\epsilon}f(R-s)\mathrm{d}s\\
&=\epsilon f(R)+O(\epsilon^2)\text{.}
\end{split}$$
Therefore
$$\begin{split}\frac{p(\epsilon)}{q(\epsilon)}&=\frac{\epsilon f(R) +O(\epsilon^2)}{\epsilon nR^{n-1}+O(\epsilon^2)}\\
&=\frac{f(R)}{nR^{n-1}}+O(\epsilon)\text{.}
\end{split}$$
so the limit in question is equal to
$$\begin{split}
 \lim_{ \epsilon \to 0} \frac{1}{R^n-(R-\epsilon)^n} \int_{R-\epsilon\le \|x\| \le R} \mathrm{e}^{-\frac{\|x-\mu\|^2}{2}} dx
&=\frac{f(R)}{nR^{n-1}}\\
&=\frac{1}{nR^{n-1}}\int_{S_{n-1}}\mathrm{e}^{-\tfrac{1}{2}\|R\hat{x}_{\Omega}-\mu\|^2}R^{n-1}\mathrm{d}\Omega\\
&=\frac{\mathrm{e}^{-\tfrac{1}{2}R^2-\tfrac{1}{2}\mu^2}}{n}\int_{S_{n-1}}\mathrm{e}^{\hat{x}_{\Omega}\cdot\mu R}\mathrm{d}\Omega
\end{split}$$
In turn, the remaining integral is expressible in terms of a modified Bessel function:
$$\int_{S_{n-1}}\mathrm{e}^{\hat{x}_{\Omega}\cdot a}\mathrm{d}\Omega =(\tfrac{a}{2})^{1-n/2}S_{n-1}\Gamma(\tfrac{n}{2})I_{n/2-1}(a)\text{.}$$
(Here $S_{n-1}$ is the content of the unit $(n-1)$-sphere.) Therefore
$$\lim_{ \epsilon \to 0} \frac{1}{R^n-(R-\epsilon)^n} \int_{R-\epsilon\le \|x\| \le R} \mathrm{e}^{-\frac{\|x-\mu\|^2}{2}} dx
=\frac{\mathrm{e}^{-\tfrac{1}{2}R^2-\tfrac{1}{2}\mu^2}}{n}(\tfrac{\mu R}{2})^{1-n/2}S_{n-1}\Gamma(\tfrac{n}{2})I_{n/2-1}(\mu R)\text{.}
$$
