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This is problem thirty nine of this to Rick Calculus
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Safe Edition section two point four. If the function
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F is defined by F of X is equal to
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zero, if excessive is rational, one of X
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is irrational. Prove that the limit is experts.
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Zero of the function f does not exist. We
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recall the definition of the limit on using Del Delta
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and Absalon, and here's a graphical representation of that
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definition. We see that poor given function as you
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approach X. You should approach the limit. L
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, um provided that there are there exists now use
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for Delta For every given. Absalon, for example
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, here's a closer minus epsilon tolerance for L,
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and we see that it corresponds to a certain TelDta
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tolerance for X. Now, as absolute get smaller
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, we should see that it also has a corresponding
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, smaller delta range. And so in this case
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, we would be able to prove that the limit
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for this function exists from because of this consistent epsilon
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. Delta found it parameters they were able to find
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. If we deal with this function, half let's
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plot a little bit of what this might look like
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. Affects zero If X is rational. So let's
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pick some rational values. How about well, at
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one over here it's rational, the value one half
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his rational. Maybe the value in making it a
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zero point one. Jealous, irrational. And,
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of course, Ciro. I would also be rational
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. And because these are rational points, the function
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is given a value zero breach of these points.
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Irrational numbers are a very large set of numbers,
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and in fact, there are an infinite amount of
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irrational numbers between any two rational numbers. For example
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, anywhere around here wait, what? I need
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to know specifically what that irrational number is, but
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we know that there will be some points where in
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between a couple of rational numbers, there is an
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issue irrational number, meaning that the function we'll have
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a value of one. And so what we see
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is dysfunction is definitely not continuous. It has values
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of between their own one, and it jumps very
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often. So it's very Unser. And as we
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get closer to zero, this is where we're trying
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to prove whether this exists. We see that we
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cannot define a certain Absalon tolerance for any assault tolerance
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we can not to find a certain Delta tolerance,
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which is what our definition of a limit is currently
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, Um, as we let's say, we set
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a certain Delta tolerance around here. We see that
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this I could not allow for inappropriate, absolutely tolerance
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in order for us to define this limit. Andi
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element in this case does not exist because of the
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nature of the function as it jumps around a lot
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because there are an infinite many of international numbers,
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as well as rational members to the left, into
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the right of zero. And for this, for
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these explanations in these reasonings, we say that the
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limit, his expertise here of dysfunction does not exist
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.