finding delta for limits: prove lim, x->2, x^2 equals 4

[trdert]

hi all,

i'm working on some problems at calc on the web.

The limit of x² as x approaches 2 is 4. That is,
lim _{x -> 2} x² = 4

Let epsilon= 0.2.
Find delta > 0 such that

if | x - 2 | < delta
then
| x² - 4 | < 0.2

so far i've plugged f(x), L and epsilon into f(x)-L > epsilon to get:
x²-4 < 1/5, and i can see that it makes sense to go from there to
(x-2)(x+2)<1/5

so i need to get rid of the x+2 somehow to arrive at the same format as that for delta, but i have no idea how.

i'm having similar issues with this problem:

The limit of x³ as x approaches 3 is 27. That is,
lim_{x -> 3} x³ = 27

Let epsilon = 0.2.
Find delta > 0 such that

if | x - 3 | < delta
then
| x³ - 27 | < 0.2

i can see that x-3 is a factor of x³-27 but no idea on the cancelling

[stapel_eliz]

The idea behind epsilon-delta proofs is that you can find a delta for any epsilon. So fixing the epsilon at some named numerical value misses the point, I'm afraid.

You have |x² - 4| that you're trying to show is less than 4. You want to be able to say that, for any epsilon I pick, you can give me a delta such that, if |x - 2| is less than the delta you gave me, then |x² - 4| is less than the epsilon I gave you. So let's look at the inequalities we're working with:

We start by assuming that:

. . . . .

. . . . .

. . . . .

We want to prove that:

. . . . .

The above means the following:

. . . . .

. . . . .

. . . . .

. . . . .

By nature of quadratics, the final inequality above is true for:

. . . . .

Comparing with the "delta" inequalities above, let:

. . . . .

. . . . .

Since is positive, then so is . Now see where this leads....