60–81. Limits Evaluate the following limits. Use l’Hôpital’s R...

Question

60–81. Limits Evaluate the following limits. Use l’Hôpital’s Rule when needed.

lim_x→1 ( x- 1)^sinπx

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the limit as X approaches 5 of X minus 5 raised to the power of sine pi X divided by 5, applying Lapita's rule if necessary. A says it's -1, B 0, C1 and D says the limit does not exist. Now, if we're gonna evaluate the limit before we apply Lapita's rule, let's just establish that we need the rule before we continue, OK? So let's try to evaluate the limit by direct substitution. Now, that means we're going to put 5 into our expression, OK? So by direct substitution, that means we're gonna be finding uh X the limit as X approaches 5 of our our expression here. Now, what does that mean? Well, as X approaches 5, then we know that X minus 5 is going to approach 0, OK? And on the other hand, we also know that at X equals 5. OK. Then the sign of pi X divided by 5 is gonna be equal to the sign of 5 pi divided by 5, which is the sign of pi, which equals 0. So now, as X approaches 5, then our limit. Is going to, in other words, our limit as X approaches 5 of our expression. Is going to approach 0 to the power of 0, and that is an indeterminate form for a limit, OK? So now that we have established, we know that it will end up in an indeterminate form, then we can start to apply Lopita's rule. Now How can we do that? Well, since we have the indeterminate form 0 to the power of 0, it suggests that we can take the logarithm of the expression and then apply the rule if necessary. So let's let Y, OK, let's let Y be equal to X minus 5 raise to the sign of I X divided by 5, OK. Then taking the natural logarithm of both sides, that means the natural log of Y will be equal to the natural log of X minus 5 raises to the power of the sign of pi X divided by 5. Now in this case, OK, then applying the laws of logarithms, that means we could rewrite our right hand side as the sign of pi x divided by 5 multiplied by the natural log of X minus 5. I know as we go along, it's always good to check if we can directly evaluate our limit just to ensure that we don't waste our time in using Lapita's rule, right? Now, here, in this case. If we were to substitute here, uh, as X approaches 5, OK, then, or at X equals 5, the natural log of X minus 5 would be negative infinity and the sign of pi X divided by 5 would approach 0. So that would lead to the indeterminate form of 0, OK. 0 multiplied by negative infinity, so we can still continue trying to use Lopita's rule. Now, let's rewrite our expression as a caution, OK? And now if we do that, then we can rewrite it in the form, the natural log of X minus 5, OK. Here, let me write that 5 properly. Divided by the reciprocal. Of the sign of pi X divided by 5. And no, because this is in or or written as a reciprocal, we can start to think about Lapita's rule. And Lapita's rule basically tells us that we could rewrite our limit, OK? In other words, it says that if we have a limit, let me put that on the right here. OK Lapita's rule basically tells us have some space, so let's make use of it. Basically tells us that if we have a limit. As X approaches A of FX divided by G of X. And it's indeterminate in an indeterminate form. Then we could find our limit, OK. By differentiating our numerator and our denominator. So we separately. So in other words, we could find it's the same thing as the limit as X approaches A of the derivative of F of X divided by the derivative of G of X. So if we apply Lapita's rule here, OK. Then that means we could rewrite our expression. As the limit. As x approaches 5 of the derivative of our numerator. L N X minus 5. Divided by the derivative of our denominator. The reciprocal of sine by x divided by 5. And now when we do that the derivative of our numerator. OK. Is going to be equal to 1 divided by X minus 5. Let me fix that here. Let's clean that up a little bit. OK. One divided by X minus 5. I know the derivative of our denominator, OK, is going to be, it's gonna be quite an expression here. It's gonna be negative pi divided by 5. by the cosine of pi X divided by 5. OK. Here we're differentiating it by the chain rule, all divided by the square of sine pi X divided by 5. OK. And now if we simplify our expression even further, then we should get it to be equal to the limit as x approaches 5 of negative 5 signs quad pi x divided by 5. Divided by pi multiplied by X minus 5, multiplied by the cosine of pi X divided by 5. No, let's test. Will this give us our limit? Are we able to now find our limits since we've differentiated the numerator and denominator? OK. Well, no. That means then. That we're going to find the limit, OK, as X approaches 5 here and as X approaches 5, we know that the sign of pi X divided by 5 approaches 0. So this is gonna be -5 multiplied by 0 squared, or I could write it right here. It's -5 multiplied by 0 squared. OK. And in our denominator, right, we know that the cosine of pi X divided by 5 will approach -1 and X minus 5 will approach 0. So this is gonna be pi multiplied by 5 minus 5 multiplied by -1, which is going to be negative 0 divided by 0. But now, we've run on the same problem we're trying to run away to get away from. Again, we have an indeterminate form. But no, to find or to solve, we can use Lapita's rule again to resolve this indeterminate form, OK? So now here. So again, let's just clean that up. Applying Lopita's rule again, OK, what does that mean? What does that mean? But it now means we're going to take our indeterminate form and again differentiate numerator and denominator. So in other words, no, we're going to be finding the limit. Let, let me just kind of copy this here in the interest of time. We're going to take our expression. OK, let me just clean this part. And no. All we're really saying is that if I just move this over a little bit, we're going to find the derivative of our numerator, OK, of the new numerator and the derivative of the new denominator. Now, what is that going to look like? Let me write that in blue. Well, now again we write our limit as x approaches 5, and now we know that in our denominator the derivative of -5 sin skrit pi x divided by 5 is going to be 2 pi spi x divided by 5 multiplied by the cosine of pi x divided by 5 when you differentiate it using the chain rule. In our denominator, we can differentiate this expression by using the product rule and as we do, we should get it to be equal to pi cost by X divided by 5. Minus I squared multiplied by x minus 5. Multiplied by the sine IX divided by 5. Now, all of this is divided by 5, OK? Now that we have it have it in this form again, let's try to evaluate by substituting or finding our limit through direct substitution. Now, when we substitute, OK, we should find that we get 2 pi multiplied by the sign of 5 pi divided by 5 or sign pi multiplied by the cosine of pi, essentially the same thing. All divided by pi cost pi. Minus pi squared multiplied by 5 minus 5 multiplied by the sine of pi divided by 5. And now that we have that we can evaluate what do we know? Well, no, we know, of course, that uh the sign here, the sign of pi is 0, OK? So you can write this has 2 pi multiplied by 0, and the sign of, uh sorry, the cosine of pi is -1. On the other hand, in our denominator, that means we're gonna have pi multiplied by -1 minus 0 because 5 minus 5 is 0, it turns this entire term to 0. And no, 2 multiplied by 0 multiplied by -1 is 0, negative pi minus 0 equals negative pi and 0 divided by negative pi equals 0. Now, if we recall the form that we had it in earlier, it was the natural logarithm of Y. So that means as the natural logarithm of, or sorry, our natural logarithm of Y is approaching 0, which means that Y is approaching E to the 0 or 1. Therefore, that tells us the limit as X approaches 5 of our original expression. X minus 5 to the port of signed pi X divided by 5 is going to be equal to 1. If we look back at our answer twice, OK, that means C is the correct answer. Thanks for watching everyone. I hope this video helped.

60–81. Limits Evaluate the following limits. Use l’Hôpital’s Rule when needed. lim_x→1 ( x- 1)^sinπx

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60–81. Limits Evaluate the following limits. Use l’Hôpital’s Rule when needed.
lim_x→1 ( x- 1)^sinπx