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.
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 $$\lim_{ \epsilon \to 0} \frac{1}{R^n-(R-\epsilon)^n} \int_{R-\epsilon\le \|x\| \le R} \mathrm{e}^{-\frac{\|x-\mu\|^2}{2}} dx =\frac{\mathrm{e}^{-\tfrac{1}{2}R^2-\tfrac{1}{2}\mu^2}}{n}(\tfrac{\mu R}{2})^{1-n/2}S_{n-1}\Gamma(\tfrac{n}{2})I_{n/2-1}(\mu R)\text{.} $$