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 → −∞ (2x + √(4x² + 3x − 2))

Accepted Answer

Hello. In this video, we are going to be determining the limit of the given function by using conjugation. The limit given to us is the limit as X approaches negative infinity of 3 X plus the square root of 9 X2 + 5 X minus 1. Now, in order to approach this problem, we first need to figure out what type of conjugate we are going to work with. But in order to find a conjugate, we first need to rewrite our limit in the form of a rational function. We can do this by placing the given function over 1, and we can rewrite the limit to be the limit as X approaches negative infinity of 3 X plus the square root of 9 X2 + 5 X minus 1 divided by 1. Now, what kind of conjugate are we going to multiply this quantity by? Well, whenever we see a square root quantity, the conjugate ends up being the same quantity, but with a different sign. So instead of us having a sum of two terms, the conjugate is going to be a difference of two terms. So in this case here, the conjugate of the given function is going to be 3X minus the square root of 9 X2 + 5 X minus 1. And if we multiply the numerator of this rational function by by the conjugate, we must do the same thing to the denominator. So we are multiplying the denominator by 3X minus 9 X2 plus 5 X minus 1. Now what we need to do is we need to simplify the function. In the numerator, we are going to foil these quantities together, and by foiling the quantities, we can simplify the numerator to be 9 X2 minus the quantity, 9 X2 plus 5 X minus 1. And in the denominator, if we multiply one to the conjugate, we just get the conjugate. So we have 3 X minus the square root of 9 X2 plus 5 X minus 1. Now what we need to do is we need to simplify this function. We will start by distributing -1 into the quantity 9X2 + 5 X minus 1. What that's going to do is this is going to flip the signs of each of the terms. So we will end up having negative 9 X2 minus 5 X + 1. By combining the like terms, we can cancel out the 9 x 2 terms from the numerator, and that is going to leave us with negative 5 X + 1 divided by 3X minus the square root. Of 9 X 2 plus 5 X minus 1. So, what we have here is a simplified form of the function, but this function still has the limit as X approaches negative infinity. Now, the trick here is that we want to divide everything by the highest power of X. The reason being is because we will end up with some terms that are going to be some number divided by infinity once we apply the limit, and anything divided by infinity is going to zero out those terms. So, let's go ahead and first simplify the function by factoring out some terms of X. If we take a look at the square root, we can factor out an X2d since this is the highest power of X. When we do that, we will end up with a denominator of 3 X minus the square root of X2 multiplied by the quantity, 9 + 5 divided by X minus 1 divided by X squared. And we can simplify this even further by pulling out the X term from the square root. This will allow us to simplify the square root of X2 to just be X. So we can simplify the denominator to be 3 X minus X multiplied by the square root of 9 + 5 divided by X minus 1 divided by X2. For the numerator, we can do the same thing, and we can factor out the highest power of X, which is going to be X. That will allow us to rewrite the numerator to be X multiplied by -5 plus 1 divided by X. Plug putting these back into the original limit will leave us with the limit as X approaches negative infinity of X multiplied by -5 + 1 divided by X, divided by 3 X minus X multiplied by the square root of 9 plus 5 divided by X minus 1 divided by X squared. Notice that we can factor out an X from the numerator and denominator, and this will allow us to simplify the limit even further to be the limit as X approaches negative infinity of -5 plus 1 divided by X, all divided by 3 minus the square root of 9 + 5 divided by X minus 1 divided by X squared. Now, what would happen if we apply the limit here? If we apply the limit, anything that is divided by some factor of X is going to become 0. So we are left with -5. In the numerator Divided by 3 minus the square root of 9, but there is a problem here. Square root of 9 simplifies to be 3, and 3 minus 3 gives a 0, which is undefined. So instead, we have to plug in a value of X that is approximately slightly greater than 0 in order to check the sign of the square root. Now, the square root can never have a negative number inside of it, so we will go ahead and pick a very small positive number to plug in for X. If we approximate this value, we end up getting a number for the square root that is slightly larger than 3. And once we plug in the limit, we will get -5 divided by some negative number, which is going to be negative infinity. And this is going to simplify to give us positive infinity, which is going to be the value of the limit. So I hope this video helps you in understanding on 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 → −∞ (2x + √(4x² + 3x − 2))

Key Concepts

Limits at Infinity

Conjugate Multiplication

Dominant Term Analysis

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