Is $r$ when $\lim_{r\rightarrow 0}$ "equivalent" to $\frac{1}{n}$ when $\lim_{n\rightarrow\infty}$?

The question I am trying to do is:
Let $f:U\in\mathbb{R}^n\rightarrow\mathbb{R}$ be a continuous function.
Show that for every $x\in U$ we have: $$f(x)=\lim_{r\rightarrow 0}\frac{1}{vol.(B(x,r))}\int_{B(x,r)}f(y)dy$$


I showed that the following is true: $$f(x)=\lim_{n\rightarrow\infty}\frac{1}{vol.(B(x,\frac{1}{n}))}\int_{B(x,\frac{1}{n})} f(y)dy$$
The way I did it was that given $n$, there is a point $a_n\in B(x,\frac{1}{n})$ such that the RHS$=f(a_n)$. So, we have a sequence $\{a_n\}_{n=n_1}^\infty$ that converges to $x$ ($n_1$ is the smallest natural such that $B(x,\frac{1}{n_1})\subset U$). Then, the sequence $\{f(a_n)\}_{n=n_1}^\infty$ converges to $f(x)$

My question is: are these equivalent? I can think of cases where you can't change the $r$ for $\frac{1}{n}$, for instance, some function $g:(0,1)\rightarrow\mathbb{R}$ where $g(\frac{1}{n})=0\forall n\in\mathbb{N}$ but it keeps oscilating while in between these values in a way that $lim_{r\rightarrow0}g(r)$ does not exist. However, I feel like I can do it in this question, because of the way I chose $a_n$, but I am not totally sure