Why is in the definition of a limit of a function 0

Question

Why is in the definition of a limit of a function 0<|x-p|<E, not just |x-p|<E? Also, I have a ton of textbooks, and in one of them, it actually is simply |x-p|<E. What's the difference?

amistre64 · Answer

i believeit is because you want to determine a value as distance between x and p approaches zero

amistre64 · Answer

but then again, it could be author preferences

anonymous · Answer

Could you give an example, something is a limit by second definition, but is not by the first?

amistre64 · Answer

not off hand.

but calculus was plagued at the beginning of the divide by zero scenario

so the limit concept was defined as a way to get around it; you can pick a number smaller than epsilon, that is not 0, yada yada

amistre64 · Answer

the second one looks more like it is part of an actual calculation to me ....

anonymous · Answer

It seems that by the first definition functions from isolated points (I don't know exactly how they are called in english) would not have limits. Wonder why they didn't like isolated points.

amistre64 · Answer

an isolated point has not definable slope does it?

amistre64 · Answer

waking up and typing just dont mix ... :)

anonymous · Answer

nope, but slope does not seem to be a part of the definition?

anonymous · Answer

it appears later, in derivatives, if I understood right

amistre64 · Answer

that is my thought as well

amistre64 · Answer

the value of the limit of an isolated point is the point itself, since the point itself defines its neighborhood, i recall reading someplace

amistre64 · Answer

the 0<n<E definition appears to be defining left and right limits such that the value at f(x) is moot

anonymous · Answer

Ah! Somehow, suddenly I got it. I think. Thanks!

amistre64 · Answer

good luck ;)