WEBVTT
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So for this problem, we are going to be
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discussing limits. Um, and were given the limit
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as X approaches A of, um we're just gonna
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plug in a immediately. So in the numerator we
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get on the square root of two a cube X
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minus X to the fourth, and then that's gonna
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be minus eight times the cube root of a X
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. So when we plug in a what we're gonna
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end up getting is that that's equal to the square
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root of to A to the fourth minus a to
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the fourth minus a cube times this square or the
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cube drew of a cube. So it's just eight
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of the fourth. And then similarly, we will
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end up with, um eight of the fourth.
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Here is Well, um, so as a result
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of that, we know that this is actually a
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squared minus a squared, which is zero. And
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then the denominator will have the limit as acts of
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purchase A of a minus, the the root of
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X cubed when we do that one of getting a
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minus thebe route of a cute or a to the
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fourth. So regardless of what we get here,
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we know that in plugging in a we're going to
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get zero as well. So based on this,
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what that tells us is that we are dealing with
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the indeterminant form 0/0. So we're gonna wanna use
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Low Patel's rule. When we differentiate the numerator and
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denominator, we're gonna end up with the limit as
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X approaches a, um are numerator is going to
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be when we have we take the derivative and then
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we evaluated at a, um we're gonna get the
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limit. Actually, we can already plug in the
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values, so let's get rid of this limit.
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So once we take the derivative, we can just
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plug in the values, so we're gonna get one
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half eight of the fourth to the negative one half
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time to negative to a cube minus a cubed over
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three times a cube to the negative, two thirds
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over negative 3/4 a cube age of the fourth to
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the negative three force and we can do the algebra
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here. A lot of simplification. Ultimately, what
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we're going to get down to is 12 a plus
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for a over nine, which is equal to 16
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a over nine, and that will be our final
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answer