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Problem 34 Hard Difficulty

Verify, by a geometric argument, that the largest possible choice of $ \delta $ for showing that
$ \displaystyle \lim_{x \to 3} x^2 = 9 $ is $ \delta = \sqrt{9 + \varepsilon} - 3 $.

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This is Problem thirty four of the Stuart Calculus eighth edition, Section two point four. Verify by a geometric argument that the largest possible choice of Delta for showing that the limit is X approaches three of the function X squared equals nine is Delta is equal to the square root of the quantity nine plus Epsilon minus three. So using our excellent out the definition of a limit, we begin first with our restriction for O. R. This initial assumption if the absolute value of the quantity X minus A in our case is three, that's what X is approaching. If this is less than Delta, then so we're going to rewrite these inequalities. Then the function, which is X squared minus the limit value is nine going be. Listen, Absalon. So we're going to keep working with just this first inequality. For now, this absolute value can be re written as negative Delta Less than X. Ah minus three. It's Listen, Delta Ah, And then what we can also right is that adding sixty both sides or two each term should give us relationship for X Plus three and the reason we chose ex ministry and expose story for each of these is to have ah two terms that when multiplied give us the function X squared minus nine s o. We're going to do just that. We're going to show that X squared minus nine X squared minus nine. This sub's values equivalent to, ah, the product of explain his three absolute value of that and X plus three. Since we know the range for ex ministry and expose story in terms of daughter, we're going to choose the maximum leech such that when we do choose that. So for ex ministry, the maximum is Stilton. For X plus, Terry, the maximum is dull Triple six. Then we know for sure that this is less than Absalon, which is what the initial and equality here states. So we used that restrictions. Oh, are the assumption for Delta? Ah, here. And then we wrote this second inequality in terms of Delta and absolute Ah, if we rearranged this inequality, we should get dealt the squared just by distributing doubt. The Squared plus six Stelter minus epsilon is less than zero. We're going to continue working on this right up here. If we attempt to complete this, where in order to have it have this peace terms and the more ah convenient form. Completing this Cory means that we should add nine and subtract nine and we'LL see why this is ah useful. Ah, this term here these terms here tell two squared plus six doctor plus nine can be reduced or factored into the quantity X or adult upholstery quantity squared and we see that the remaining terms are negative. Nine. Indicative Absalon Aiken Write that as my negative or minus d quantity nine plus Epsilon and for ah convenience again to have a convenient form for this inequality we treat. We'LL treat this as a difference of squares so that we can oh, factor even further. The square here is dull tapestry. That's what's being squared here. We can say the Delta are the quantity that's being squared is the square root of this quantity. So we're going to have a doctor plus three. So this is the first term. Yeah, minus what is being squared over here, which is the square root of nine plus epsilon. And then the other term is the same. But just plus doctor three Wes square root of nine minus a nine plus Epsilon, and this is all supposed to be less than zero came. Now we look at each of the terms. It's two terms are multiplied to become less than zero. One will have to be positive. We won't have to be negative. We can see that the second term here has to be positive because we stated in the definition of limit in the absolute Delta format. Absolutely, though they're always positive numbers to a positive value doctor here, plus tree plus square root of nine plus another positive number. This is a positive. That must mean that this first term is exclusively negative dot upholstery minus the square root of nine plus Absalon is less than zero. That's what negative means, and we can go out and sell for. Delta must be less than, um, the square root. I have nine plus scene plus Epsilon minus three, and it's at this point that we can see it's at this point that we can see that Delta can be any number that's greater than zero, but less than this quantity, depending on what Absalon is chosen. If that's the case, this is the maximum value. So we stayed at the maximum Delta town is equal to this value. Nine plus absolutely. This quantity in the square root squared of Naples absent minus three. And this is, Ah, our final answer and we have proven this statement.

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Problem

(a) For the limit $ \displaystyle \lim_{x \to 1}(…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Kristen Karbon

Samuel Hannah

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Kristen Karbon

Samuel Hannah

James StewartCalculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Kristen Karbon

Samuel Hannah

James Stewart

Calculus: Early Transcendentals