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Problem 24 Medium Difficulty

Find the limit or show that it does not exist.

$ \displaystyle \lim_{x \to -\infty}\frac{\sqrt{1 + 4x^6}}{2 - x^3} $

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2

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Video Transcript

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