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Let's say I have a function $x^2$ and $x_o=2$, Then by Cauchy definition: $f:Df \rightarrow R$ is continuous at a point $x_o=2 \iff \forall \epsilon>0 \ \ \exists \delta>0: x\in(Df \cap V_{\delta}(x_o)\setminus(x_0)) \Rightarrow |f(x)-f(x_0)| $

Given $\epsilon$ does there exist $\delta>0: x\in(Df\cap 0<|x-2|<\delta) \Rightarrow |x^2-4|<\epsilon $

Does there exists $\delta>0: x\in (Df \cap 0<|x-2|<\delta) \Rightarrow |x-2||x+2|<\epsilon?$ If we choose $\delta\le1$ Then it follows that if $x\in(Df \cap 0<|x-2|\delta \Rightarrow |x-2|\cdot5<\epsilon \iff |x-2|<\frac{\epsilon}{5} $ Then if we choose $\delta=min(1,\frac{\epsilon}{5})$ we get that If $x\in (Df \cap 0 <|x-2|\le \delta \Rightarrow |x^2-4|\le\epsilon$

So, I basically did it the way a limit is proved for a function at a certain point, I don't know whether that is correct and if you can prove that f is continuous at a point choosing just one particular $\delta$ if it must be true for all possible $\epsilon$?

Also how it can be proved using Heine's definition of continuity?

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Partial answer. According to Cauchy definition:
Given $\epsilon >0 $ there is $\delta>0: x\in(Df\cap |x-2|<\delta) \Rightarrow |x^2-4|<\epsilon$.

We have that $2-\delta<x<2+\delta$ or $x+2<4+\delta$. Thus, $|x^2-4|=|(x-2)(x+2)|<\delta(4+\delta)$. So we need to pick $\delta$ to satisfy $\delta (4+\delta) < \epsilon$.

Solving the last inequality we get $0< \delta <\sqrt{4+\epsilon}-2$. Thus, such $\delta$ exist and the function is continuous at $x=2$.

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