Please forgive me if this question has a trivial answer, but I’m a bit stuck after coming up with this scenario. If there is a function with a point/removable discontinuity, for example $f(x)=1$ when $x\neq 0$ but $f(x)=0$ when $x=0$. The limit as $x$ approaches $0$ is $1$ but with the epsilon delta definition, if we take $\epsilon = 0.1$, then if $x=0$ (which is in the range $|x-a|<\delta$), $|f(x)-1|>0.1$, which is the negation of the epsilon delta statement. What exactly am I missing?
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The $\epsilon-\delta$ definition for continuity is different form the $\epsilon-\delta$ definition for limit, notably for continuity the value $x=x_0$ is allowed but not for the limit for which $x\neq x_0$.
user
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Oh wow! I didn’t know that. Is there somewhere, implicit or explicit, in the precise definition for real valued functions (ex. Wikipedia that says that’s the case? I just thought it had to be in the domain for the limit definition. – rb612 Mar 23 '18 at 11:19
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@rb612 yes of course for the definition of continuity $x_0$ have to be in the domain but it is not requested that it is necessarly a cluster point, indeed a function is continuos also in an isolated point. – user Mar 23 '18 at 11:22
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@rb612 you can find other related OP on this topic on MSE as https://math.stackexchange.com/questions/776159/understanding-epsilon-delta-continuity-definition?rq=1 – user Mar 23 '18 at 11:24
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@rb612 https://math.stackexchange.com/questions/15616/continuity-of-a-function-at-an-isolated-point – user Mar 23 '18 at 11:25
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What makes the definition correct (limit being defined as) if there is no explicit mention that $x$ cannot equal $x_0$? I’ve never seen that restriction stated in the definition. – rb612 Mar 23 '18 at 11:27
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@rb612 for the limit we assume $0<|x-a|<\delta$ which means $x\neq a$ – user Mar 23 '18 at 11:28
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oh! That’s what I missed! Thank you! – rb612 Mar 23 '18 at 11:33
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@rb612 see also the related https://math.stackexchange.com/questions/2628911/why-are-we-allowed-to-cancel-fractions-in-limits/2628914#2628914 – user Mar 23 '18 at 11:34
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The original question was posted long time ago. However, I had the same doubt. I think I understand it a little better after some research.
So the Epsilon-delta definition states we can find a delta around x to satisfy that if |x-a|< delta then |f(x)-L|< any small number (Epsilon). Here f(x) doesn't necessarily need to be evaluated at "a". The definition such state that f(x) evaluated at all values at |x-a|< delta. Meaning delta is always >0 and cannot be equal to zero.
here's a nice video explaining that: https://www.youtube.com/watch?v=0sCttufU-jQ
