WEBVTT
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So we're trying to determine the limit as X is
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approaching infinity of the function one plus four X to
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the sixth. All over two-X. Cute.
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Now if we were to do direct substitution, you
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would end up with a square root of one plus
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four times infinity to the sixth power over to minus
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infinity to the third power which ends up being infinity
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over infinity, which is what we refer to as
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an indeterminate form. So that means there's some other
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way of solving this. So what we're gonna do
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is we're going to take that problem and we're going
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to multiply by a form of one And that form
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of one is going to be one over the square
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root of X to the 6th. In doing so
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I'm going to share that with the numerator and with
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the denominator. So we will now have the limit
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as X is approaching infinity. Uh the square root
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of one plus four X to the sixth. Over
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the square root of X to the sixth. And
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that in turn is all over two minus X cubed
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over the square root of X to the sixth.
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So in the top, since they both have the
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same radical, we could rewrite that as one radical
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one plus four X to the sixth over X to
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the sixth. And in the denominator we could share
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that denominator with each part of the numerator and say
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to over the square root of X to the 6th
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minus X cubed over the square root of X to
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the sixth. In the numerator. We could now
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share that denominator and say the limit as X approaches
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infinity of the square root of one over X to
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the 6th Plus four X to the 6th over X
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to the 6th. And the denominator we're just going
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to keep as is Let's do a little bit more
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clean up and then we have to talk about the
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square root of X to the 6th. So in
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the numerator we can clean up a little bit further
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And we could say the square root of one over
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X to the 6th plus four because these factors would
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in turn cancel. And in the denominator we now
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have two over the square root of X to the
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6th minus X cubed over the square root of X
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to the sixth. Now let's chat about the square
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root of X to the 6th. The Square Root
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of X to the 6th is going to be an
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X cubed. But we also have to employ the
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absolute value of X cubed. And when we talk
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in terms of absolute value then we could say that
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this is equivalent to X cubed as long as X
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is greater than or equal to zero but it's equivalent
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to the opposite of X cubed when X is less
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than zero. And since we are letting X approach
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a positive infinity we're going to use the top part
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because our X when it's approaching a positive infinity,
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X would be greater than or equal to zero.
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So in place of both of the square root of
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X to the 6th power we can put a plain
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old x cubed. So we now have the limit
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as X is approaching infinity of the square root of
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one over X. to the 6th plus four over
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two over x cubed minus X cubed over, execute
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Yeah. And if we clean up one more time
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, the limit as X approaches infinity of the square
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root of one over X to the sixth plus four
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over two over X cubed minus one. So we're
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now ready to do direct substitution. So if we
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were to substitute our infinity in place of the two
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variables, we would now have the square root of
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one over Infinity to the 6th power plus four.
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All over to over infinity cubed minus one. And
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any time we're dividing by a super large number that
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is approaching zero. So we would end up with
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0-plus four. Again here we were dividing by a
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super large number so we'd have 0-1 which leads
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us to the square root of four over negative one
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or two over negative one or negative too. So
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if we go back to our original problem, the
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limit as X approaches infinity of that expression is going
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to be negative two. Now we can also reinforce
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that. Looking at the calculator. We if I'm
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going to bring in my graphing calculator in my y
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equals. You can see that I have placed the
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expression in. And if I look at my table
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And I put numbers in, so notice I've placed
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in 200 and I got a negative 2.5. As
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I put in 2000, I got negative two.
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As I put in uh two million, I got
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negative, too, so I can reinforce it in
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my calculator.