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 → −∞ (√(x² + 3) + x)

Accepted Answer

Calculate the limit of the function F of X equals square root of 9 X2 + 4 minus 3 X as X approaches negative infinity and X approaches positive infinity. Where we have 4 possible answers, limits that approach either 0 or infinity. Now to solve this, let's first modify our function. F of X equals the square root F9X2 + 4 minus 3 X. We'll multiply this by the conjugate to simplify it a little better. In this case, the conjugate will be the square root of 9 X2 + 4 + 3 X. Will also put the same thing in the denominator of a fraction. Wound up getting The square root of 9 X2 + 4 minus 3X. Multiplied By the square root of 92 + 4 + 3 X. Divided by the square root of 9 X2 + 4 + 3 X. This simplifies to be 9 X2 + 4. Minus 9 x2 all divided by the square root of 9 x2 + 4. Plus 3 X. Was further simplifies to 4. Divided by the square root of 9 X2 + 4 + 3 X. Now that we're here, we can take our limit. The limit. As X approaches infinity, we'll go first. Of 4, divided by the square root of 9 X2 + 4. Plus 3 X. Now, we can simplify this. By multiplying the radical by X and changing the inside part of the radical. 29 + 4 divided by X squared. If we factor X2d out from everything inside the radical, and then take the square root of that, we end up getting the square root of 9 + 4 divided by X2 remaining in the radical. We'll add on the 3 X and then factor out. The X's and the denominator. We end up getting 4 divided by X multiplied by the square root of 9 plus 4 divided by X2 + 3. Now, we can rewrite our limit. We have the limit, as X approaches infinity. F 4 divided by X, multiplied. one Divided by the square root, 9 + 4 divided by X2 + 3. Which can be further rewritten as the limit as X approaches infinity of 4 divided by X multiplied by the limit. As X approaches infinity. Of 1 divided by the square root, 9 + 4 divided by X2 + 3. We can then calculate the limit. We get 0 multiplied. By one Divided by the square root. 9 + 4 divided by infinity plus 3. We end up getting 0, multiplied by 1/6, which is 0. So the limit As X approaches infinity, is equals to 0. Now, let's set the limit for negative infinity. To limit X approaches negative infinity. Or Divided by the square root. 9 X2 + 4 + 3 X. For this one, let's take a look at the denominator. So we have the square root of 9 X2 + 4. A square root of x2 + 4 is approximately 3 absolute value of X, which breaks down to 3 X. And -3 X. Now, we have negative infinity, so we want the negative form of this approximation. Meaning we have Or divided by. -3 X plus 3X, because the square root of 9 X squared. Plus 4 is approximately -3 X. We get 4 divided by 0. This means the limit, as X approaches negative infinity, this equals 2 positive infinity. We now have both limits for our answer. As it approaches positive infinity, our answer is 0, and as it approaches negative infinity, our answer is positive infinity. If we look at our possible answers, we see that the answer to our problem is answer A. OK, hope to help you solve the problem. Thank you for watching. Goodbye.

Finding Limits of Differences When x → ±∞

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

lim x → −∞ (√(x² + 3) + x)

Key Concepts

Limits at Infinity

Conjugates in Limits

Square Root Functions

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