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Prove the statement using the $ \varepsilon $, $ …

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Problem 31 Hard Difficulty

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to -2} (x^2 - 1) = 3 $

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

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Question 2: Why is there a possibility of delta to be one why not just one fifth epsilon

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Question 1: Why did you replace the absolute value of x-2 by 5.

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Video Transcript

Mrs Problem number thirty one of this tour calculus eighth edition section two point four Prove the statement using the absolute out the definition of a limit. The limit is exports is negative. Two of the quantity X squared one is one is equal to three. The Epsilon Delta definition of Lim states that this limit is equal to three as experts negative too. Four. And provided that for any absalon greater than zero you confined the delta greater than zero such that if the absolute value of the difference between X and A is less than delta in the absolute value of the difference between the function f of X and l, the limit is less than epsilon. Our first step will be to use the second inequality and plug in our given function. F which is X squared minutes one subtracted slim it which is three said this Liston Absalon simplifying we get steps. The value of the quantity X squared minus four is less than absolute fact dream. Since it's a difference of squares, we get the absolutely of the quantity X plus two times X minus two. Listen, Absalon and we can separate each of the terms separately into their own absolute values. And we pause here momentarily to notice that we cannot yet find and inequality similar to this first inequality that we we note. Since we always want to make a connection between Absalon and Delta, this inequality here will be explained. It's a A B negative, too, so X minus things to his ex, plus two less than Delta. So we cannot yet make a comparison between the two s. So it's important that we make an assumption for this quantity absolutely of X plus two on DH. See if we can determine the relationship between Delta and excellent in that way, since we are approaching native to and we're very close to negative two, it's a fair assumption to say that this inequality is less than one to tell you. Within a small distance. This statement is the same as saying that negative one is Liston. X plus two is less than one. I'm not subjecting four from each term we get negative. Five is Liston X minus two is less than, ah, negative three. Since he's objected for from each term now, the reason we did this is we want to have a comparison of this quantity explains to hear that we found in this blue last line for inequality we're going to make a small change here to this inequality since explains to his lesson into three. It definitely will be less than positive five. And we choose that number because then now we can transform this inequality into the absolute value of X minus two is less than the value five for convenience because we know that this comparison exists now, based on this first initial assumption, we go back in here, we replace explains to with five and we're going to find an Absalon based on this condition. On this assumption, we get that that's the value of X. Plus two is less than the less than absolute divided by five. And now we have three statements for ex lap civilian extras too. This one relates it. That quantity with Delta, this one relates to Absalon. And this was their initial assumption. No. So we stayed to Delta Must be either one or absolute over five. However, since we want to satisfy both of these here Ah, these ah inequalities. We have to ensure that we choose the minimum, choosing the minimum of either of these terms, depending on what Absalon is guarantees that we satisfy all of the conditions laid out by the definition of the limit based on steps on Delta definition and this if we follow this choice of Delta, we are able to prove that statement as such.

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