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Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\lim _{x \rightarrow 1} \frac{x^{a}-a x+a-1}{(x-1)^{2}}$

$\frac{a(a-1)}{2}$

Calculus 1 / AB

Chapter 4

Applications of Derivatives

Section 3

L'Hospital's Rule: Comparing Rates of Growth

Differentiation

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All right. So for this problem we are evaluating the limit is X approaches one of X to the power of a minus a X plus a minus one over X minus one squared. So first, we will evaluate the limit. X equals one, and we'll get one thio power of a minus a plus a minus one over one, minus one squared. And so one of the power A no matter what a is will be equal to one. So we end up with 0/0, everything cancels out and so we can apply, Lupita. Lt's rule. That means this would be equivalent to the limit as X approaches one, and we'll take the derivative of the expression in the numerator. So it's going to be a Times X to the power of a minus one minus a. And of course, a and minus one are both Constance with respect to X. So those are both going to just become zero, and then we'll take the derivative of the denominator. And so this is going to be two times X minus one. Okay, so now that we have taken the derivative, we can evaluate the limit again, and put one in for acts to suit. We gets. That's gonna give us a times one to the power of a minus one minus a over two times one minus one. Okay, so no matter what, one to the power of a mess one is This is gonna end up being equal to a minus a and then we have two times zero, which, of course, is gonna be 0/0, so it can use local calls. Rule a second time, and we'll get that. This is equivalent to the limit as X, the approaches one of. And so I want to take the derivative of this first expression eight times X to the power of a minus one, and that will give us a times a minus one times X, The power of a minus two and then this minus A has no exes in it. Right? So we're taking the derivative with respect to X, so that's gonna become zero. So we're done with the numerator. We'll take the driven it with respect to X for the denominator then. And what we're left with is a two. So the two X would become the two and then the minus two. Here, it will become zero. So we will evaluate again. We'll put one in for acts and see what we get. We're gonna have a times a minus one times one to the power of a minus. Two over too. Okay, well, so this one is just gonna stay a one, right? No matter what it is. So what we get is a times a minus one over to then this is going to be the solution for the limit.

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