Let $f ∶ E \subset \mathbb R^n$ and $x_0 \in \mathbb R^n$. Show that $$\lim_{x \to x_0}f(x) = l$$ $\iff$ $\exists R > 0$ and a function $g ∶ ]0, R[ \to \mathbb R^+$ such that $\lim_{r \to 0^+}g(r)=0$ and for all $x \in B(x_0,R) \cap E/ \{x_0\}$, $|f(x) − l| \leq g(\lVert x − x_0 \rVert)$.
For the $\Longrightarrow$ implication, I have $\forall \epsilon > 0, \exists \delta > 0$ such that $0 < \lVert x- x_0\rVert < \delta \implies \lvert f(x)-l \rvert < \epsilon$. I took $r$ to be defined as $r:= \lVert x-x_0 \rVert$ so that when $r$ goes to $0$. But then I don't really see what can be my function $g$ (I was given the hint that it should involve a sup so I thought of $g(r) := (\sup {\lvert f(x) -l\rvert}, x \in B(x_0,r) \cap E/\{x_0\})$ but I don't really understand why then what would be R ?
For the $\Longleftarrow$ implication, I take $r:= \lVert x-x_0 \rVert$ and I have $\forall \epsilon > 0, \exists \delta > 0$ such that $0 < r < \delta \implies \lvert f(x)-l \rvert < g(r) <\epsilon$ then $\lim_{x \to x_0}f(x) = l$.
Can someone help me for the first implication ?
I am thinking that since $E$ is an open neighborhood around $x_0$ (not necessarily all of $\mathbb R$), take $R$ to be smaller than the radius of any open ball around $x_0$ contained in $E$ to guarantee $g$ is well-defined. This should ensure $f$ is bounded on the ball.
If you take $R$ to be too large, $\sup$ may go to infinity. For example consider $f(x)=1/x, x_0=1, R=2$ and see what happens with your $g$.