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I know the two definitions for continuity, (sequential and epsilon-delta)

Given $x_0 \in D, \forall \epsilon > 0, \exists \delta > 0, |x - x_0| < \delta \rightarrow |f(x) - f(x_0)| < \epsilon $

and

f is continuous if $\forall x_n \rightarrow x_0 $ implies $f(x_n) \rightarrow f(x_0)$

Now the definition of $\lim_{x \rightarrow x_0} f(x) = L$ is $$\forall \epsilon > 0, \exists \delta > 0 |x - x_0| < \delta \rightarrow |f(x)-L| < \epsilon$$

So are the two $\epsilon-\delta$ definitions the same? ie $\lim_{x \rightarrow x_0} f(x) = L$ implies $f$ is continuous at $x_0$? I want to make sure.

FFjet
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MinYoung Kim
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    If you assume that $x_0$ is in the domain of definition of $f$, then the answer is yes. – amsmath Oct 06 '19 at 02:01
  • I posted an answer relatively recently about this. – Theo Bendit Oct 06 '19 at 02:06
  • That is not the definition for $\lim_{x\mapsto x_0} f(x) = L$. It should read $0 < |x-x_0|<\delta$. And that's kind of the point here. – Torsten Schoeneberg Oct 06 '19 at 02:17
  • People, some of the answers and comments here are confusing. Please check with the function $f(x) = \begin{cases} 0 \text{ if } x \neq 0 \1 \text{ if } x=0 \end{cases}$. For $x_0 := 0$, obviously $x_0 \in D$, obviously $f$ is not continuous at $x_0$, obviously $\lim_{x\to x_0} f(x) = 0$, although with the wrong definition in the OP, $\lim_{x\to x_0} f(x)$ does not exist. – Torsten Schoeneberg Oct 06 '19 at 02:48

2 Answers2

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$\displaystyle \lim_{x \to x_0} f(x) = L$ implies $f$ is continuous at $x_0$ if and only if $L=f(x_0)$.

azif00
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You are given that $$ \forall\epsilon>0\,\exists\delta>0 : |x-x_0|<\delta\,\Longrightarrow|f(x)-L|<\epsilon. $$ For any $\epsilon>0$ you can thus choose $x=x_0$ (because $|x-x_0|=0$) to get $|f(x_0)-L|<\epsilon$ and thus $f(x_0) = L$. Hence, you can replace $L$ above by $f(x_0)$ to see that $f$ is indeed continuous at $x_0$.

amsmath
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