# Homework Help: Question on the construction of a limit

1. Oct 19, 2007

### Simfish

http://www.math.caltech.edu/classes/ma1a/Fa07Ma1aHW2.pdf [Broken]
Problem 2

(note, I'm not a Caltech student, so this is not a violation of the Honor Code - also, solutions have been posted)
The solution to problem 2 is at
http://www.math.caltech.edu/classes/ma1a/07Ma1aSol2.pdf [Broken]

The question with the solution, how is this epsilon constructed? It seems to come out of nowhere. I tried constructing an epsilon myself.

This is what I tried:
"the question I asked was "how in the hell did the person get N? Clearly, one can do it by expressing n in terms of epsilon, but if you do that, you merely get n = (epsilon+1)^1/3 / (1 - (epsilon + 1)^1/3). of course, this > 1 / (1 - (epsilon + 1)^1/3. (and the (epsilon + 1) should not be dif. from the (epsilon - 1) in computations since the inequality reverses (and that should give you an constant dependent on epsilon that n must be bigger than.

So is the only step you have to take to express n in terms of epsilon, and then to find a value of n that is strictly less than that to ensure that the limit of the function is always less than epsilon as n -> infinity?

The question is - is the constant of the official solution better than the constant I gave?

Last edited by a moderator: May 3, 2017
2. Oct 19, 2007

### mjsd

this is classic example how to do proof of this type using "definition of a limit". the key as you have pointed out is to get N in terms of epsilon in such way that the desire inequality can be proved. once you have identified a N (may be as a function epsilon), then that means you can always choose N (using that rule) so that the inequality you are trying to prove is correct.

put it simply, the fact that you are able to choose N as a function of epsilon like this (and the inequality desired holds) means the limit actually exist, and by demonstrating that such choice exist you have proved the problem (ie. like proof by example). Alternatively, if you can't find such a choice (and there are cases like that), that means the limit does not exist. Sometimes, it is however harder to prove that.

the issue about n vs N: by the definition of the limit, all you need is find such n large enough, it doesn't matter how large/small, it just has to be large enough and finite.