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Verify that another possible choice of $ \delta $ for showing that $ \displaystyle \lim_{x \to 3} x^2 = 9 $ in Example 4 is $ \delta = \min\{2, \varepsilon/8\} $.

$\left|x^{2}-9\right|<\varepsilon$

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

Limits

Derivatives

Campbell University

Oregon State University

Harvey Mudd College

Boston College

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this is problem number thirty three of the story calculus eighth edition, Section two point four. Verify that another possible choice of Delta for showing that the limited exit purchased three of the function X squared is equal to nine. An example for is Dr Equals minimum are too and absolute or eight. So recall that the abscond off the definition of a limit is that if the difference between X and a the absolutely of the differences less than Delta Tau, then the absolutely the difference between the function and the limit is Absalon less than Epsilon, provided that Absolute and Delta are both greater than zero. So we begin and just used this definition directly and use our daughter values. Not seeing if axe Dobbs value of X minus a three is less than we're going to use this first. Also assuming that this is the minimum. Listen to, then. If this is true, we can rewrite it as a negative, too less than explain his three at less than two and where we're going to do is we're going to take ah, we're going to commit this to X plus three instead in the middle and in order to do this, it means adding six to each term. Six minus two or six plus negative too is for six plus two eight. And in order to modify this and change back to Napster value, we see that expose trees. Lesson eight experts three is also greater than four Mommy extra stories also greater than any value. Listen for and for comedians will choose negativity. And we do this because now we can easily see that this is the value of quantity. Extras three is less in it. Okay, now that we have this term, we turned to our second option for Delta and that is the absolute value. X minus two Sorry, X minus three is less than ah Absalon rate. And what we see with these two at terms is that if we multiply the two interplay next, plus three, the quantity extreme and multiplying the corner he explain this three it is the same as the absolute value of the quantity experts Tree terms Ex Ministry, which turns out after foiling to be equal to X squared minus nine. Multiplying these two terms eyes the same as multiplying these two terms. And if we want to buy these two terms, Absalon over eight, not abide by it. We see that the council, meaning that it's illegal to just excellent on this right side of this equation. So what we have in the end is that the absolute value of the function X squared when it's time is less than Epsilon. And we see that this is very similar to this condition for FX, where the absolutely of the function X squared is we see his F minus to limit, which is nine is less than epsilon. So we have proven our statement. We verified that this new delta this different before my delta also confirms our limit to pay some the absolute out the definition of a limit.

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