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

Limits, differentiation, related rates, integration, trig integrals, etc.
trdert
Posts: 6
Joined: Tue Aug 04, 2009 1:21 pm
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### finding delta for limits: prove lim, x->2, x^2 equals 4

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

stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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:

. . . . .$|x\, -\, 2|\, <\, \delta$

. . . . .$-\delta\, <\, x\, -\, 2\, <\, \delta$

. . . . .$2\, -\, \delta\, <\, x\, <\, 2\, +\, \delta$

We want to prove that:

. . . . .$|x^2\, -\, 4|\, <\, \epsilon$

The above means the following:

. . . . .$-\epsilon\, <\, x^2\, -\, 4\, <\, \epsilon$

. . . . .$4\, -\, \epsilon\, <\, x^2\, <\, 4\, +\, \epsilon$

. . . . .$0\, \leq\, x^2\, <\, 4\, +\, \epsilon$

. . . . .$x^2\, -\, (4\, +\, \epsilon)\, <\, 0$

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

. . . . .$-\sqrt{4\, +\, \epsilon}\, <\, x\, <\, \sqrt{4\, +\, \epsilon}$

Comparing with the "delta" inequalities above, let:

. . . . .$2\, +\, \delta\, <\, \sqrt{4\, +\, \epsilon}$

. . . . .$\delta\, <\, \sqrt{4\, +\, \epsilon}\, -\, 2$

Since $\epsilon$ is positive, then so is $\delta$. Now see where this leads....