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When trying to take the limit as X goes to infinity of a function that involves a square
root, especially a fraction that also has a square root, we're still going to do the
same kind of technique we did with rational functions -- trying to divide through by the
highest power of X in the bottom. But there's going to be a few complications on how this
works. So we're going to do an example and see how it works. So we have the limit as
X goes to infinity of 3 X cubed divided by the square root of 9 X to the sixth plus X,
and we want to divide by the highest power of X in the bottom. So you might be thinking
we want to divide by 1 over X to the sixth, but that's not correct. Actually, the highest
power of X in the bottom is the square root of X to the sixth, which would be X to the
third. So, we want to divide top and bottom by X to the third. What we get when we do
that is limit as X goes to infinity of 3 X cubed over X cubed divided by 1 over X cubed
square root of 9 X to the sixth plus X. And now how we simplify the numerator algebraically
is pretty straightforward, we're just going to cancel these X cubeds. But on the denominator,
it's kind of difficult to see how we would combine these algebraically. And the way we're
going to do that is to remember the rule that the square root of A times the square root
of B is equal to the square root of A times B. So we need to rewrite 1 over X cubed as
something inside a square root. And if we want to know what do we take the square root
of to get 1 over X cubed, it's 1 over X to the sixth. So we rewrite our problem as the
limit as X goes to infinity of 3 divided by the square root of 1 over X to the sixth times
the square root of 9 X to the sixth plus X. Now we can combine the two square roots. So
we have the limit as X goes to infinity of 3 divided by the square root of 9 X to the
sixth over X to the sixth plus X over X to the sixth, dividing through by X to the sixth.
Simplifying that algebraically, we get the limit as X goes to infinity of 3 divided by
the square root of 9 plus 1 over X to the fifth. Now, using our limit laws, we can see
that this is going to have the form of 3 over the square root of 9 plus 1 over infinity.
And we know 1 over infinity goes to zero. So this is just 3 over the square root of
9, or three thirds, or 1.