Choose $r = R- \lvert \mathbf x - \mathbf y \rvert$ show that $B_r(\mathbf x) \subset B_R(\mathbf y)$ and compute limit

Question

I am asked to show that the ball $B_r( \mathbf x) \subset B_R(\mathbf y)$ and then to compute compute$\lim\limits_{R \to \infty} \frac{\lvert B_R(\mathbf y) \rvert}{\lvert B_r(\mathbf x) \rvert}$

We have that $\mathbf x, \mathbf y \in \Re^n,\quad R > r>0$

I am given that $r = R- \lvert \mathbf x - \mathbf y \rvert$.

I am more struggling with the first part of the question rather than the second, as I think that as $R \to \infty$ then the limit $\to 1$.

I know that for $B_r( \mathbf x)$ this means that for a point $a$ in this set we get that $\lvert \mathbf x - a \rvert < r$ for all $a$, this give then that $\lvert \mathbf x - a \rvert + \lvert \mathbf x - \mathbf y \rvert < R$.

And I also know that $B_R(\mathbf y)$ means that for a point $b$ in this set gives us that $\lvert \mathbf y - b \rvert < R$, but from this point I am not to sure what to do.

Answer 1

Let $a \in B_r(x)$. Then $|x-a| < r$, and by the triangle inequality $$|y-a| \le |x-y| + |x-a| < |x-y| + r = R.$$ Thus $a \in B_R(y)$, giving you $B_r(x) \subset B_R(y)$.