Limit as $x\to 0$ the same if $x$ restricted to be inside a ball? Here $f:\mathbb{R}^n\to\mathbb{R}$ is a function, and $B(0, R)\subset \mathbb{R}^n$ is the ball of radius $R > 0$ centered at the origin.

Is the limit of $f$ as $x\to 0$ always the same as the limit as $x$ goes to zero but when it is restricted to be in a ball of fixed radious $R$ around $0$?
$$
\lim_{x\to 0} f(x) \overset{?}{=} \lim_{\substack{x\to 0 \\ x\in B(0, R)}} f(x)
$$

I never see this mentioned anywhere.
 A: Here is Randall's comment written out more explicitly:
$a = \lim_{x\to 0} f(x)$ says: for all $\varepsilon > 0$, there exists some $\delta > 0$ such that $|x| < \delta$ implies $|f(x) - a| < \varepsilon$.
So pick some $\varepsilon > 0$. Then there exists a corresponding $\delta > 0$. Set $\delta' := \min\{ R, \delta \}$. Then $\delta' \leq \delta$, so $|x| < \delta'$ still implies $|f(x) - a| < \varepsilon$. But now we also know all $|x| < \delta'$ satisfy $x \in B(0,R)$, since $\delta' \le R$. This shows the two limits are the same.
A: The potential issue could be that with a certain norm, you may only consider some limits along a path, e.g. along $x_2=mx_1$ for instance and have an existing path limit but not a global limit.
But in $\mathbb R^n$ all norms are equivalent, so the coordinate-wise $x_i\to 0$ convergence is the same as convergence in norm $\lVert x\rVert\to 0$ and the limit is independent of the ball chosen.
Said in a different manner, you can always get an isotropic ball $\lVert x\rVert_\infty<\delta\implies \lVert x\rVert<R$ for a given norm which take care about limits along a path.
