Finding Limits of Differences When x → ±∞
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Question

Finding Limits of Differences When x → ±∞

Find the limits in Exercises 84–90. (Hint: Try multiplying and dividing by the conjugate.)

lim x → ∞ (√(9x² − x) − 3x)

Accepted Answer

Hello. In this video, we are going to be calculating the limits for the function F of X is equal to the square root of 16 X2 minus 3X minus 4 X as X approaches negative infinity and X approaches infinity. Now, in order to take the infinite limits, the first thing we want to do is you want to simplify the given square root function. In order to do this, we are going to use the method of conjugates. So, first, what we are going to do is we're going to represent the given function as a fraction. We can do this by writing 16 X2 minus 3X minus 4 X divided by 1. Now we need to ask ourselves, what is going to be the conjugate to this function? Well, whenever we have a square root quantity, the conjugate is usually a similar quantity, but with an opposite sign. Since we are taking the difference of two quantities here, that means the conjugate is going to be the square root of 16 X quad minus 3X plus 4X, and we are going to multiply this conjugate to both the numerator and denominator. Now, we will need to foil the numerators together, and that is going to leave us with 16 X squared minus 3X minus 16 X squad. And in the denominator, we are left with this square root of 16 X quad minus 3X plus 4 X. What we are going to do now is we are going to simplify the function. Now, in the numerator, adding the 16 X qua quantities together are going to cancel each other out. Furthermore, we can factor out an X squad from the square root, so we can rewrite the function to be -3 X divided by the square root of X2 multiplied by the quantity, 16 minus 3 divided by X. Then we are going to add 4 X. We can pull out the X squared from the square root, and that will leave us with -3 X divided by X multiplied by the square root of 16 minus 3 divided by X plus 4 X. Notice that we can factor out an X from the denominator, and we will have a factor of X in both the numerator and denominator. Once we factor out these these X's, we will have -3 divided by the square root of 16 minus 3 divided by X plus 4. This is going to be the simplest form of the function. Now what we want to do is we want to take the two limits. We want to take the limit as X approaches negative infinity and positive infinity. So, let's go ahead and start with positive infinity. We will take our function and set the limit as X approaches infinity. So we have the limit as X approaches infinity of -3, divided by the square root of 16 minus 3 divided by X plus 4. Now, since this function is in its simplest form, we're going to just directly apply our limits. Now what would happen with 3 divided by infinity? Well, there is a property that we have where any number divided by positive and negative infinity will produce 0. So if we directly substitute X approaching infinity for our limit, that is going to cancel the 3 minus X quantity. So we are left with -3 divided by the square root of 16 + 4. That is going to further simplify to be -3 divided by 4 + 4, which is going to give us -3 divided by 8, and that is going to be the limit as X approaches positive infinity. Now we are going to do the same thing, but with negative infinity. We will take the limit as X approaches negative infinity of -3, divided by the square root of 16 minus 3 divided by X + 4. Now, since the only difference here is that we're plugging in negative infinity, negative infinity is still going to cause 3 divided by X to be 0, and that means that the output for negative infinity is going to be the same. It will be -3 divided by 8, and these are going to be the solutions to the problem. So I hope this video helps you in understanding how to approach this problem, and I will go ahead and see you all in the next video.

Finding Limits of Differences When x → ±∞

Find the limits in Exercises 84–90. (Hint: Try multiplying and dividing by the conjugate.)

lim x → ∞ (√(9x² − x) − 3x)

Key Concepts

Limits at Infinity

Conjugate Method

Dominant Terms

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