Is this limit equal to 1? Let $x\in R^{n}$ be fixed ($x\neq 0$) and $r>0$. Evaluate the limit:$$\lim_{r\rightarrow \infty}\frac{V(B(x,r)\cap B(0,r))}{V(B(0,r))}$$
where $V$ stands for volume and $B(x,r)$ is the ball with center x and radius $r$.
Thanks in advance.
 A: The volume of $B(x,r)$ is $ar^n$ for some constant $a$ which dependends on $n$, but whose exact value will not concern us.
Let $d=||x||$ be the distance between the two centers $0$ and $x$. If $r>d$ then the triangular inequality implies that $$B_{r-d}(0)\subset B(x,r)\cap B(0,r)$$
Hence 
$${V(B(x,r)\cap B(0,r))\over V(B(0,r))}>{V(B_{r-d}(0))\over V(B(0,r))}={a(r-d)^n \over ar^n}=\Bigl(1-{d\over r}\Bigr)^n$$
When $r\to\infty$ then the last expression will $\to 1$.
A: Yes if we work with euclidian norm. Let $I(r):=V(B(x,r)\cap B(0,r))$. Then 
$$\left|\frac{I(r)}{|B(0,r)|}-1\right|=\frac 1{|B(0,r)|}\int_{\Bbb R^n}\chi_{\{||x-y|>r\}}\chi_{\{y\leq |r\}}dy.$$
In the set on which we integrate, we have $|x|²-2\langle x,y\rangle+|y|^2\geq r^2\geq |y|^2$ hence $2\langle x,y\rangle\leq |x|^2$. This is a bounded set, hence the integral is uniformly bounded by an universal constant (independent of $r$). By the way, it gives a speed of convergence.
A: All Euclidean balls have the same volume, so we may as well use $V(B(\frac{x}{2},r))$ in the denominator.  Now draw a picture to see that: 
$$V\bigg(B\bigg(\frac{x}{2},r-\frac{|x|}{2}\bigg)\bigg)\leq V(B(x,r)\cap B(0,r))\leq V\bigg(B\bigg(\frac{x}{2},\sqrt{r^2 - \frac{|x|^2}{4}}\bigg)\bigg).$$
(enzotib points out below that it's easier to use a second radius of $r + |x|/2 > \sqrt{r^2 - |x|^2/4}$.)
Since the volume of a ball of radius $r$ is $Cr^n$, $|x|$ is fixed, and the radii of both the first and last balls are order $1$ in $r$, their volumes go as $Cr^n$.  So when we divide through by $Cr^n$, we have that both the first and last expression go to $1$, forcing the middle to go to $1$ as desired.
A: Yes. Use zooming $x\mapsto \displaystyle\frac1r\cdot x$. Then volume is shrinked by $\displaystyle\frac1{r^n}$, both in the enumenator and dominator, so the quotient is fix, $0$ stays, $r$ is constantly 1, and $x$ goes to $x/r \to 0$.
