Formal Definition of a Limit and Arbitrary Values

[Deinosuchus383]

Hi,

I noticed that for certain limits (such as x^2), we supposed to state that the epsilon or delta should be less than a certain value. Is this just as a starting point?
For instance, in Thomas' Calculus, this equation asks us to assume that epsilon is less than 4 in order for the delta to apply for larger epsilons.

\(\displaystyle 0\, <\, \bigl|\, x^2\, -\, 4\,\bigr|\, <\, \epsilon\)

\(\displaystyle -\epsilon\, <\, x^2\, -\, 4\, <\, \epsilon\)

\(\displaystyle 4\, -\, \epsilon\, <\, x^2\, <\, \epsilon\, +\, 4\)

\(\displaystyle \sqrt{4\, -\, \epsilon\, }\, <\, x\, <\, \sqrt{\epsilon\, +\, 4\,}\)

\(\displaystyle \sqrt{4\, -\, \epsilon\,}\, -\, 2\, <\, x\, -\, 2\, <\, \sqrt{\epsilon\, +\, 4\,}\, -\, 2\)

I've mainly been reading Calculus wikibooks and Thomas Calculus to try to understand this.

http://en.wikibooks.org/wiki/Calculus/Choosing_delta

 

[stapel]

I noticed that for certain limits (such as x^2), we supposed to state that the epsilon or delta should be less than a certain value. Is this just as a starting point?

For instance, in Thomas' Calculus, [the proof that the limit, as x goes to 2, of x^2 is 4] asks us to assume that epsilon is less than 4 in order for the delta to apply for larger epsilons.

As the explanation states, the proof is divided into the two cases because of the specifics of the arithmetic. If epsilon is greater than or equal to 4, then, as the page at the link notes, the proof is simple. But you need to proof the limit for all cases, and especially as epsilon gets very small. When epsilon gets closer to zero, complications arise which can be avoided, or at least simplified, if one can make assumptions about the possible values of epsilon.

For instance, if epsilon is less than 4, then \(\displaystyle 4\, -\, \epsilon\) is positive, so the square root is well-defined.