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This is problem number eight of the Steward Calculus eighth
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edition section two point four For the limit shown.
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Limited's expert is one of the quantity. Either the
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two X minus one over r O I X is
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equal to to illustrate definition to pipeline values of Delta
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that correspond to absolute is equal to zero point five
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. An absolute is equal to zero point one.
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So definition to states that this limit is true as
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long as we can find a delta greater than zero
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for every Absalon created zero satisfying. Ah, In
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this case, X minus zero absolutely must be listing
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Delta value. This is true, then the function
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ah must differ from its limit by listening absolute value
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. So we're going to find it out. The
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value for each of these have some values. Two
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confirm this limit came. So this is our FX
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just for reference. We're going to rearrange the second
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inequality as they get of Absalon less than the function
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. F minus two is greater than is less than
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epsilon. We're gonna have to teach term native excellent
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plus two. Listen, if is less than absolute
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list too, and we're going to be in with
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the first absolute value of zero point five. So
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that gives us one point. Five is less than
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F is less than two point five. That's our
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first range for it for this first absolute tips on
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one, and we use a graphing calculator graphing tool
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to plot this function around this region for F is
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around to want one factor two point five, and
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we find the corresponding X values so as if we
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trace the function either to experience one over X once
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it reaches a value of about two point five.
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That is when the X values around zero point two
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. And if we trace the body down to around
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where the value is, where the y values one
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point five. We see that the excise approximately negative
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zero point three on this case because the ah a
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value zero So explain zero. We already have it
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in a form that we can compare the deltas.
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This's tell, too, and Delta Delta must be
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less than zero point to lessen or equal to,
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and must be less than or equal to zero point
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three. So we always choose a smaller of the
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two. So our first Delta Delta one, We're
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going to choose zero point two for the first Absalon
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. The second absolutely zero point one. Our range
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becomes one point nine. Listen, have is less
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than two point one and the X values that correspond
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to these f of X values R, is there
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a point zero five and negative zero point zero five
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approximately. So we see that the doubt the value
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must be less than or equal to the smaller the
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two. In our case, we're going to choose
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approximately zero point zero five, which is enough to
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confirm and validate the second absolute value. And then
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our two answers are shown here, down below.