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This is problem number forty four of the Stuart Calculus
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. Eighth edition, section two point three. Final
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amid. If it exists, it's the limit does
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not exist. Explain why Here we have a limited
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six bridges. Negative too. Of the function to
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minus work. The absolute value backs divide where the
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quantity two plus sex. Okay, since this is
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a complete limit, not just a limit from the
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left or the right, we wantto determine first.
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Whether approach in a tutu makes this an indeterminate form
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or an undefined value. Two minus the absolutely of
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negative too is to minus two two plus native who
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? Zero. So we have a zero division here
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, divide by zero. Add some*** too.
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So we know that the value or the function is
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on to find a native to so we were unable
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to determine that it exists are at this point.
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Um, but we can still prove that ism it
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exists as long as the limited left negative two equals
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limit from the right. I'm negative too. On
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that the equal each other. So what we want
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to do is we want to first deal with this
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absolute value function, and we want to rewrite this
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Lim. I'm using what we know about the absolute
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value function of X. We know that when X
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is greater than zero or equal to zero, it's
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equal the function X. However, when X is
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negative, he also value is meant to change the
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sign and so dysfunctions represented by a negative X and
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for our interests since the limit of expertise and negative
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too. This is on ly in this region.
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So we will not be considering the absolute value function
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too. Take this form because we are only concerned
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with the X approaches negative to region. So we
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will substitute dysfunction with negative X instead. So rewriting
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this limit as experts needed to two minus native X
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or plus X divided by two plus X Now notice
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that I didn't write experts in ecstasy from the left
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or expert unit two from the right because both of
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those give you the same results right since we're pushing
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native to which is a value in this region.
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So the function looks the same, whether it's from
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the left or the right. As we can see
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, we can cancel these too, and this simplifies
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to one And if we take the Limited's experts think
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, too, of this constant value of one.
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Her answer for the limit is one again keeping in
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mind that this is true for both. Pushing is
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equal to negative to pushing that from the left and
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from the right. And since those Walter True and
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exist and they both equal one, then this limit
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from the beginning also equals one, and that is
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our final answer.