Limits by power series Use Taylor series to evaluate the follo...

Question

Limits by power series Use Taylor series to evaluate the following limits.

lim ₙ → ₄ ln (x - 3)/(x² - 16)

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the following limit using the Taylor series. The limit as Y approaches to of the natural log of Y minus 1 divided by Y2 minus 4. A says it's 0. B says it's a 1/2, C says it's 1/4, and D says it's 1/8. Now, let's first think about our limit. What would happen if we try to evaluate it by substituting, or sorry, try to evaluate it as Y approaches too? Well, as Y approaches to. Then we'd substitute y equals to into the expression and in our numerator it would become the natural log. Of 2 minus 1 divided by 2 square minus 4. Not in our numerator and that will be the natural log of 1 which is 0, and in our denominator that will be 4 minus 4, which is 0. So our limit is in the indeterminate form 0 divided by 0. So because of that, we'll need to find a workaround. How can we do that? Well, let's let K be equal to Y minus 2, which implicitly implies that Y is gonna be equal to K + 2. And, and let's also say. That as why. Approaches to. OK, as you notice here, then K is going to approach 0, OK, because that is because K equals Y minus 2. If we were to rewrite the expression in terms of key, then using this substitution. Then The LN Y minus 1 divided by Y2 minus 4 could be rewritten as LN of 2 + K. -1 Divided by 2 + k all squared. 4. No, 2 minus 1 is 1, so this is LNK plus 1. And in our numerator, that's going to be uh well, let me rewrite this as 1 plus K and you see why in a minute. But that's LN 1 plus K divided by, we expand 2 + K squared, so 4 + 4K plus K 2 minus 4. And now, notice here that we, 4 minus 4 is 0. So that will leave us with the expression LN 1 plus K divided by 4K plus K squared. Now, we can expand LN1 + K using the Taylor series around K equals 0, OK? So using the Taylor series. Then really We can say that LN 1 plus K. Is approximately equal to k minus a half of k2 plus a k cubed divided by 3 minus and it goes on. So now when we substitute the Taylor expansion into the limit. OK. Then our limit now becomes the limit as K approaches 0 of K minus 1/2 of K squared plus 3 of K cubed and so on, uh, minus and so on. OK. All divided by 4K plus K2. Now if we factork from the numerator and the denominator, then we'll have our limit. We write it below it here. We'll have our limit as k approaches 0 of k multiplied by 1 minus 1/5 of k plus 1/3 of k squared and so on. All divided by K multiplied by 4 plus k. And in this case, if we cancel out the term K, then we'll have the limit as K approaches 0 of 1 minus 1 half of, of K rather. Yeah, plus. A third of K squared and so on, all divided by 4 plus K. No, as k equals 0. Ask approaches 0, we can substitute k equals 0. So when we do that, then no, notice that in our denominator this will be 4 + 0, and in our numerator as k approaches 0, k divided by 2, and k square divided by 3, and all of the terms that come after and have a k term in there would also approach 0. Thus, Our limit is going to be equal to 1/4. So now we've been able to evaluate our limit using our Taylor series. Thanks a lot for watching everyone. This, oh, my apologies, that means that C. and how could I forget? That means that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Limits by power series Use Taylor series to evaluate the following limits.

lim ₙ → ₄ ln (x - 3)/(x² - 16)

Key Concepts

Taylor Series Expansion

Limit Evaluation Using Series

Handling Indeterminate Forms

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