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

Limits, differentiation, related rates, integration, trig integrals, etc.

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

Postby trdert on Thu Aug 06, 2009 4:05 pm

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,
limx -> 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
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Postby stapel_eliz on Thu Aug 06, 2009 5:15 pm

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. :oops:

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.... :wink:
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