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imram.math
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please i am new to math. I donot know exact meanings of epsilon-delta definition. i dont comprehend it. Would Anybody help me. thanks in advance
please i am new to math. I donot know exact meanings of epsilon-delta definition. i dont comprehend it. Would Anybody help me. thanks in advance
There's no reason to exclude the point ##x = x_0##. We trivially have ##|f(x_0) - f(x_0)| < \epsilon## for any ##\epsilon##.##0<|x-x_0|<\delta## actually.
There's no reason to exclude the point ##x = x_0##. We trivially have ##|f(x_0) - f(x_0)| < \epsilon## for any ##\epsilon##.
It sure is a reason.
Otherwise, discontinuous functions would be deprived of limit values at the point of discontinuity. Thus, the limit concept would be conflated with the continuity concept.
Think about it!
Hmm..no read it again.But vanhees71 was giving the definition of continuity at ##x_0##, not the definition of the existence of a limit at ##x_0##. If the function is to be continuous at ##x_0##, then it must be defined at ##x_0## and have the correct value!
One more remark and then I'll shut up. :tongue: Here's a discussion from Spivak's Calculus (chapter 6, before Theorem 2) which may clarify:Hmm..no read it again.
What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
In particular, he writes d>0
Spivak said:If we translate the equation ##\lim_{x \rightarrow a} f(x) = f(a)## according to the definition of limits, we obtain:
For every ##\epsilon > 0## there is a ##\delta > 0## such that, for all ##x##, if ##0 < |x - a| < \delta##, then ##|f(x) - f(a)| < \epsilon##.
But in this case, where the limit is ##f(a)##, the phrase ##0 < |x-a| < \delta## may be changed to the simpler condition ##|x-a| < \delta##, since if ##x = a## it is certainly true that ##|f(x) - f(a)| < \epsilon##.