- #1

- 679

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I believe it does. If ##f(x)## is continuous then ##\lim_{x\to c}f(x)## exists. But if ##f(x)## is undefined for some values of ##x## in the ##\delta##-neighbourhood of ##c##, then we cannot say ##|f(x) - L|<\epsilon## for ##0<|x-c|<\delta##, contradicting that ##\lim_{x\to c}f(x)## exists. So there must exist a ##\delta>0## such that ##f(x)## is defined for all values of ##x## in the ##\delta##-neighbourhood.

https://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit

If so, does this mean that it is enough to just say "suppose ##f(x)## is continuous in an open subset ##U## of ##\mathbb{R}##" instead of "suppose ##f(x)##

**exists**and is continuous in an open subset ##U## of ##\mathbb{R}##"? (which is what is stated below)

http://calculus.subwiki.org/wiki/Clairaut's_theorem_on_equality_of_mixed_partials