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.

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 $\to 1$.

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.