## anonymous one year ago For the limit lim x → 2 (x3 − 4x + 3) = 3 illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.)

1. anonymous

I know how to do parablas but not cubic

2. zepdrix

I don't remember these so well :) So take what I say with a grain of salt lol When your x is within delta of 2,$\large\rm 0\lt|x-2|\lt\delta$Then the distance between your function and the limiting value should be within epsilon,$\large\rm |x^3-4x+3-3|\lt\epsilon$So you've gotta do some work on this thing, ya?$\large\rm |x^3-4x|\lt\epsilon$$\large\rm |x(x^2-4)|\lt\epsilon$And then maybe apply Difference of Squares, ya? :o

3. zepdrix

$\large\rm |x(x-2)(x+2)|\lt\epsilon$And then umm... how do we wrap these up? I'm trying to remember haha

4. anonymous

umm your guess is better than mine lol. im teaching myself this

5. zepdrix

Let's put a restriction on delta, let it be small... say less than or equal to 1.$\large\rm 0\lt|x-2|\lt\delta\le1$$\large\rm |x-2|\lt1$$\large\rm -1\lt x-2\lt 1$$\large\rm -2\lt x\lt 2$So that ummm... If x is between 2 and -2, the junk in our epsilon equation is between these values,$\large\rm 0\le|x(x+2)|\le8$The larger of those is 8, so we'll use that.$\large\rm |x(x+2)|~\cdot~|(x-2)|\lt\epsilon$$\large\rm |x(x+2)|~\cdot~|(x-2)|\lt8~\cdot~|(x-2)|\lt\epsilon$ So then uhhh... we would want the minimum value for this distance (x-2). Either $$\large\rm \frac{\epsilon}{8}$$ or $$\large\rm \delta$$.

6. zepdrix

Hopefully I'm doing this right :P Look at that, lemme know if it doesn't make sense..

7. anonymous

Im confuse I think im going to a math lab thanks for your help

8. zepdrix

Woops, when I added the 2 to each side of the inequality it should have given this: $$\large\rm 1\lt x\lt 3$$ Which would change things a bit. But ya, go to math lab c: figure some stuff out.