Limits at a real number, using epsilon and delta

[Nerpilis]

OK I probably have some dumb questions here but it might be partially due to the lack of examples at my disposal and minimal explanation in my text.
\lim_{x\to{0}}(x+1)^{3} = 1

\mid f(x) - L \mid < \epsilon
\mid(x+1)^{3} - 1\mid < \epsilon

now I know that delta is as follows:
0 < \mid x - a\mid < \delta
0 < \mid x - 0 \mid < \delta
as far as i know that i need to pick delta = min{1, ??}. I know that you pick 1 for convention but from my example i' am little stumped on how to get the other delta choice. If anyone knows of some other links to examples of these types of problems please send them along.

 

[Jeff Ford]

Usually you try to find your delta as a function of your epsilon. In this case I'd try to get \vert (x+1)^3 -1 \vert into a form where it was \vert x-0 \vert multiplied by something, then solve for \vert x-0 \vert in terms of \epsilon
Then you've got a relationship between \delta and \epsilon

 

[shmoe]

The basic plan is to factor |f(x)-L| into a bounded part and a part that you can control easily by picking an appropriate \delta. That is get |f(x)-L|=|x-a||g(x)|. You know this will be bounded by \delta B, where B is a bound for g(x).

To find this B it's usually necessary to have restricted the values of x first by assuming an upper bound of delta. If your assume \delta\leq 1 then you know x is on the interval (a-1,a+1), so you can bound g(x) according to this. 1 won't always work, you need to make sure that g(x) is bounded on this restricted interval.

You can then find delta in terms of epsilon and this bound B.

Take a look at http://www.math.toronto.edu/~joel/137/handouts/limits.pdf for some more examples.

 

[Nerpilis]

ok good hint i think so far:
\mid(x+1)^{3} - 1\mid < \epsilon multiplied out then extracting the factor x:
\mid x \mid \mid x^{2} + 3x + 1 \mid < \epsilon
\mid x \mid < \frac {\epsilon}{\mid x^{2} + 3x + 1 \mid}
now this is in terms of what delta is greater than. At this point I am assuming i use the L =1 for the x obtaining :
\mid x \mid < \frac {\epsilon}{5} thus choosing

\delta = \min {1, \frac{\epsilon}{5} }

 

[Jeff Ford]

Not quite. You need to control \mid x^{2} + 3x + 1 \mid} before you divide \epsilon by it.

This is where the assumption that \delta <= 1 comes in.

Set \vert x-a \vert < 1 and then use that to define a range for x. Use that to define a range for \mid x^{2} + 3x + 1 \mid}, then use the maximum value from that range for x.