Can two functions with the same limit everywhere disagree at uncountably many points? As the title states, I am curious for two functions $f,g:\mathbb{R}^d\rightarrow\mathbb{R}^d$ that disagree at uncountably many points but have the same limit everywhere, or a proof that this cannot be the case. It is clear that there are a number of examples of functions whose limits agree everywhere but disagree at countably infinitely many points, such as $$f(x)=\begin{cases}1 & x\in\mathbb{Z}\\0&\text{otherwise}\end{cases}\;\text{ and }\;g(x)=0,$$ but I can not think of how this could be true if they disagree at uncountably many points. However, I am struggling to prove this, so I'm unsure.
Thanks in advance for any help. 
 A: It's impossible. Let $f,g \colon \mathbb{R}^d \to \mathbb{R}^d$ two functions such that $D = \{ x : f(x) \neq g(x)\}$ is uncountable. For a positive integer $n$, let
$$D_n = \{ x : \lVert f(x) - g(x)\rVert \geqslant 1/n\}\,.$$
Then there is an $n$ such that $D_n$ is uncountable (for $D = \bigcup D_n$). Since $\mathbb{R}^d$ is second countable, $D_n$ has a limit point (actually, many), call it $p$. Then if $f$ and $g$ both have a limit at $p$, these limits must be different.
A: By "the same limit everywhere" I assume you mean that the limits exist everywhere and are equal.
Suppose $B = \{x: f(x) \ne g(x)\}$ is uncountable.  Consider the
 sets $B(a,b,r) = \{x: |f(x) - a| \le r, |g(x) - b| \le r\}$ where $a, b \in \mathbb Q^d$, $r \in \mathbb Q$, and $|a-b| > 2r$.  Every member of $B$ is in some $B(a,b,r)$, and there are only countably many $B(a,b,r)$, so some $B(a,b,r)$ must be uncountable.  An uncountable subset of $\mathbb R^d$ must have a limit point.  If $c$ is a limit point of $B(a,b,r)$, say $x_n \to c$ with $x_n \in B(a,b,r)$, then $$\left|\lim_{x \to c} f(x) - \lim_{x \to c} g(x)\right| \ge |a - b| - 2 r > 0$$
