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