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I could only show that a function is continuous if a $x_{0}$ was given, let's say given function is $f(x)=x^{2}$ and I need to show continuity at $x_{0}=0$.

I could do that but what if someone told me to show it for all $x_{0}$?

Is a task like that possible at all?

Same question I could ask for differentiability, what if someone told me for all $x_{0}$ or not give me any $x_{0}$?

I have been asking this myself for a long time now and I'd really like to have an answer to that question and appreciate your help very much.

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    $\begingroup$ Let $x_0$ denote an arbitrary element of $\mathbb R$ and prove that $f$ is continuous at $x_0$. If that's done then you are ready. Same for differentiation. $\endgroup$ – drhab Aug 10 '16 at 10:57
  • $\begingroup$ drhab is that correct? Ok let's say task is: Show that $f(x)=x^{2}$ is continuous in every $x_{0}$. Then I do: $\lim_{x\rightarrow x_{0}^{-}}(x^{2})=x^{2}$ and $\lim_{x\rightarrow x_{0}^{+}}(x^{2})=x^{2}$ both same thus function is continuous in every $x_{0}$? $\endgroup$ – tenepolis Aug 10 '16 at 10:59
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    $\begingroup$ Almost. the RHS of the equalities in your comment must be $x_0^2$. $\endgroup$ – drhab Aug 10 '16 at 11:01
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You show it (in symbols) for an arbitrary $x_0$, then you note that since $x_0$ was arbitrary, it holds everywhere. That's the basic procedure to show something holds everywhere.

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  • $\begingroup$ Ok let's say task is: Show that $f(x)=x^{2}$ is continuous in every $x_{0}$. Then I do: $\lim_{x\rightarrow x_{0}^{-}}(x^{2})=x^{2}$ and $\lim_{x\rightarrow x_{0}^{+}}(x^{2})=x^{2}$ both same thus function is continuous in every $x_{0}$? $\endgroup$ – tenepolis Aug 10 '16 at 10:59
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    $\begingroup$ Yes, except the limit is equal to $x_0^2$. $\endgroup$ – Sanderr Aug 10 '16 at 11:02
  • $\begingroup$ Ok thank you, very easy then. $\endgroup$ – tenepolis Aug 10 '16 at 11:02
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We need to prove that for every $\varepsilon >0$ exists $\delta>0$ so that implication holds for every $x$: $|x-0|<\delta \implies |x^2-0|<\varepsilon$. This really holds for $\delta=\sqrt{\varepsilon}$, so function is continuous at $x=0$.

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