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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!
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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##.