Question

I need some help interpreting an example problem using the \(\epsilon-\delta\) limit definition.

Answer 1

Prove that \[\lim_{x \rightarrow 2}x^{2} = 4\] I know that you end up with \(|x – 2||x + 2| < \epsilon\) and that \(0 < |x – 2| < \delta\). Now, they say that for all x in the interval (1, 3), you know that |x + 2| <5. Then, letting \(\delta\) be the minimum of \(\frac{\epsilon}{5}\) and 1, it follows that, whenever 0 < |x – 2| < \(\delta\), you have \[|x^2 – 4| = |x – 2||x + 2| < (\frac{\epsilon}{5})(5) = \epsilon\]Now, I get everything up to there they say that x in the interval (1, 3) is true to |x + 2| < 5. I don't get anything afterwards. Can someone explain to me?

Answer 2

Actually, I don't get the interval part either.

Answer 3

can we start all over again?

Answer 4

Alright.

Answer 5

Ok, now you can start looking to the difinition of the limit , vwhich's \[\lim_{x \rightarrow x0} f(x)=L\]

Answer 6

This is an example problem, so they just gave me the prove that \[\lim_{x \rightarrow 2}x^2 = 4\] using the \(\epsilon-\delta\) limit definition.

Answer 7

now you probably know the two inequalities that we should form to find delta in temrs of epsilon, which are |x-2|<delta , and |\[x^{2}\] -4 | <epsilon

Answer 8

Yes, that is what I am up to in my thinking process. I just can't figure out how to show that the \(\delta\) is possible.

Answer 9

you just need to give at least one possible \[\delta\] , usually we find the maximum delta

Answer 10

if one \[\delta\] is possible then the limit exists

Answer 11

can I go on ?

Answer 12

Yes. That's what I meant with my statement above if I was unclear.

Answer 13

sorry it's just my English lol,,
anyways, now we try to get rid of the absolute value in both inequalities

Answer 14

we get -epsilon< \[x^{2}\] < epsilon ,, and ,, -delta< x-2< delta

Answer 15

sorry it's epsilon<x^2-4< epsilon

Answer 16

I feel im doing bad so far :(

Answer 17

Ok, so you just split the absolute value into double inequalities right? No, you're doing well. :) Originally, I was trying to get where they got something from in the solution, but if there is a way of figuring it out this way, then it is great!

Answer 18

aha you are right , so we now try to figure that out dude,
just solve for x in both inequalities...
we have: 4-epsilon <x^2 < 4+epsilon ,then, \[\sqrt{4-\epsilon}< \left| x \right| < \sqrt{4+\epsilon} \]

Answer 19

now since x goes to a positive number which 's 2 , then we get \[\sqrt{4-\epsilon}< x < \sqrt{\epsilon+4}\]

Answer 20

you can from this point consider two cases which are :
1) epsilon > 4 and i this case we can't use the current inequality we just come back to the one before taking root's .
2) epsilon <4 and in this case we use it ...
am I clear so far

Answer 21

I get up to where you found the roots. I still don't quite get why you needed to find those roots in the first place.

Answer 22

to find two intervals where x belongs to,I mean one in terms of delta and the other in terms of epsilon so that we can make the endpoints equal to each other,, and choose the smallest delta between the two deltas I got after making the two endpoints equal to each other

Answer 23

so the smallest delta I take between the two ,, we call it the maximum possible delta

Answer 24

Ok, I think I understand now.

Answer 25

OK, that's great,, now considering the (second case) , the inequalities I got are \[\sqrt{4-\epsilon} < x <\sqrt{4+\epsilon}\] and \[2-\delta < x < 2+\delta \]

Answer 26

so take \[2-\delta = \sqrt{4-\epsilon} \] and \[2+\delta =\sqrt{4+\epsilon} \]

Answer 27

I think now you can solve for delta :)

Answer 28

So, then delta would be solved in terms of epsilon?

Answer 29

yes, and you can say that delta = minimum of {2-sqrt(4-epsilon) , sqrt(4+epsilon)-2}

Answer 30

remember that if only one delta exists , then the limit exists.
and in our case there are infinite number of deltas

Answer 31

Ok. Thank you. I'm a beginning calc student in self-study, so it's a bit hard to learn the material.

Answer 32

dude don't forget about the first case lol

Answer 33

self study is really helpful ,, keep up do not let anything let you down ,, I just finished calculus in the summer sission

Answer 34

If I asked you to solve the first case on your own, will that be possibility ?

Answer 35

Possibly. Just takes a while when they stump me like this.

Answer 36

But I feel like I've taken up enough time lol. I'm sure once I take a break and come back to the problem, I can figure it out.

Answer 37

okay , I will just give you a hint , before taking roots,, 4-epsilon is a negative number so the inequality will be automatically held ..

Answer 38

I hope I was helpful for you, even though I believe I wasted alot of time while typing D:
anyways If you couldn't do it just send me a msg or whatever u like
thank u for your pationce

Answer 39

Thank you for the help. :)