Let $E \subset \mathbb{R}^k, f: E \to \mathbb{R^m} $ be a function. Let $x_0 \in \mathbb{R}^k$ such that $x_0 \notin E$ but note that $ \displaystyle \lim_{x \to x_0}_{x \in E}=y_0$ does exist. Consider the function: \begin{align}g: \begin{cases} E \cup \lbrace x_0 \rbrace & \longrightarrow \mathbb{R}^m \\ x & \longmapsto \begin{cases} f(x), & x \in E \\ y_0, & x =x_0 \end{cases} \end{cases} \end{align} and show that it is continuous at $x_0$
Before I try to tackle the actual problem I want to highlight that my Professor used the following definition of the limit of a function (which is flawed in my opinion) but might be necessary to solve this problem.
Let $E \subset \mathbb{R}^k, F \subset \mathbb{R}^k$ with the function $f:E \to F$. Let $x_0 \in \mathbb{R}^k$ and $a \in \mathbb{R}^m$
$ \displaystyle \lim_{x \to x_0}f(x)=a \iff \forall \epsilon > 0, \exists \delta >0, \forall x \in E, (d(x,x_0)< \delta \implies d(f(x),a)< \epsilon )$
This definition does not exclude that $x=x_0$ should not be possible, I'd much rather use the definition
$\displaystyle \lim_{x \to x_0} f(x)=a \iff \forall \epsilon > 0, \exists \delta > 0, \forall x \in E, (0<d(x,x_0) < \delta \implies d(f(x),a)< \epsilon )$
which highlights that the function never really gets evaluated at $x=x_0$
My approach:
(1) Let $x \in E$ such that $g(x)=f(x)$. Since the limit $\displaystyle \lim_{x \to x_0}f(x)=y_0$ does exist I can say that (using the more precise definition) $ \forall \epsilon > 0 , \exists \delta > 0, \forall x \in E, (0< d(x,x_0)< \delta \implies d(f(x),y_0)< \epsilon )$ since $g(x)=f(x)$ I can also say that $d(g(x),y_0)< \epsilon$ but now I don't see how to make the connection between $y_0$ and $x_0$ to complete the proof that the function is continuous at $x_0$
(2) Let $x=x_0$ such that $g(x)=g(x_0)=y_0= \displaystyle\lim_{x \to x_0} f(x)$ but considering the definition of the existent limit $\displaystyle \lim_{x \to x_0} f(x)=a \iff \forall \epsilon > 0, \exists \delta > 0, \forall x \in E, (0<\underbrace{d(x,x_0)}_{0, \text{ contradiction?}} < \delta \implies d(f(x),y_0)< \epsilon ) $
Maybe I am applying the definition wrong but I wouldn't see how I could work with this, if I would apply the definition my Professor introduced I could say that: $$\displaystyle \lim_{x \to x_0} f(x)=a \iff \forall \epsilon > 0, \exists \delta > 0, \forall x \in E, (\underbrace{d(x,x_0)}_0 < \delta \implies d(f(x),y_0)< \epsilon )$$ and since this distance would always be true $\forall \delta >0 \implies d(f(x),g(x_0))< \epsilon$
$ \implies d(f(x_0),g(x_0))< \epsilon $ at which point I am again stuck.
I would appreciate some hints or insight whether or not my approach is correct and your opinion about which definition I should use, does it make any difference at all for this problem?
Update 2:
For (1)
$d(g(x),y_0)< \epsilon \implies \displaystyle \lim_{x \to x_0}g(x)=y_0=g(x_0)$ if that condition holds, which I have big doubts about, that would show that $g$ is continuous at $x_0$, but it somehow also implies that $x \in E$ and at the same time $x \notin E$. It is easy to see from the above equations that $f(x)=g(x)$ as $x \to x_0$. As $x$ approaches $x_0$ the functions $g$ and $f$ behave the same and both share the same limit $y_0$. However I get stuck there (yet again), I am trying to end up with an expression such like $\displaystyle \lim_{x \to x_0} g(x)=g(x_0) \implies g$ is continuous at $x_0$