Cosine limits Let n be a positive integer. Evaluate the follow...

Question

Cosine limits Let n be a positive integer. Evaluate the following limits.

lim_x→0 (1 - cos xⁿ) / x²ⁿ

Accepted Answer

Welcome back, everyone. In this problem, we want to find the limit as X approaches 0 of 1 minus 2X cubed divided by X to the 6th. A says it's -1, B 0, C1, and the D says it's 2. No, the easiest way for us to find this limit is by trying to directly substitute, OK. So let's substitute our limit or substitute our value of 0 into our limit, OK? So no, when we do that 1 minus + 2 x cubed divided by x to the 6th, it means that anywhere we see X we're going to put 0. So now this is going to be 1 minus the cosine of 2 multiplied by 0 cubed. All divided by 0 to the 6th. No 2 multiplied by 0 cubed is 0 and the cosine of 0 is 11 minus 1 equals 0. On the other hand, 0 to 6 equals 0. Now, we've run into a bit of a problem here because we're trying to divide by 0, and that means our limit is in an indeterminate form, OK, because we cannot divide by 0. So what do we do from here? Well, no, it would be appropriate to apply Lapita's rule. Recall that Lapita's rule basically tells us that if we're finding the limit as X approaches some value A of a quotient F of X divided by G of X and it's in an indeterminate form, then we can rewrite our limit, OK, as X approaches this value. Of the derivative of the numerator divided by the derivative of the denominator of our quotient, so let's differentiate our numerator and denominator in our limit, and then let's see if we'll be able to find the limit and it will not lead to an indeterminate form. So now that means we can let FFX. Be equal to 1 minus the cosine of 2Xcude, OK. And we can let G of X or denominator be equal to X to the 6th. Now, when we differentiate our numerator, OK, then we're going to get 6 X squared. Multiplied by the sign of 2 X cubed, OK. And for a denominator, its derivative is going to be equal to 6 X to the 5th. Now that we have our derivatives, then we can apply Lapita's rule. Which means then that our limit. Is not going to be 6 X squads 2 X to the 3rd. Divided by 6 X to the 5th. And we're finding this limit as X approaches 0. OK, so let me rewrite or a 0 here. No, we could simplify our expression a bit more, OK, because notice here that we can cancel 6 from the denominator and numerator and we can divide through by X2 or sorry, we can cancel out X2d from our numerator and denominator. And now this is gonna leave us with the limit as X approaches 0 of the sign of 2 X cubed, OK. Divided by execute. Now, if you notice here, the argument of our assign function is a bit similar to our denominator. And in this, we can apply the standard limit. Recall that for assign function. The limit as. Z approaches 0 or the sign of Z divided by Z, OK, where Z just represents the argument of the sign and the denominator, it's going to be equal to 1. In other words, if the argument and the of a sign function and its denominator is the same and we're finding the limit as it approaches 0, it equals 1. So the question is, can we write our expression here to make X cubed 2 X cubed so that it will be equal to the argument of sine? Well, if we try to let our argument, our Z be equal to 2XQ, OK, then notice here that we could rewrite our limit. Yes so the limit here as X approaches 0. Of a side of 2 excubed divided by excubed. We could rewrite this as the limit as x approaches 0 of 2 multiplied by the sign of 2 x cubed divided by 2X cubed. In other words, we could multiply our numerator and denominator by 2. No, when we have it in this case, the argument is the same as the denominator. So we could rewrite this as 2 multiplied by the limit as X approaches 0. Of the sign of twixt cubed divided by twixt cubed. And now based on the standard limit, this limit is gonna be equal to 1. So this is gonna be 2 multiplied by 1, which equals 2. Thus, 2 is our correct answer. If we look back at our answer choices, that means D is the right option. Thanks a lot for watching everyone. I hope this video helped.

Cosine limits Let n be a positive integer. Evaluate the following limits.

lim_x→0 (1 - cos xⁿ) / x²ⁿ

Key Concepts

Limit of a Function

Taylor Series Expansion

L'Hôpital's Rule

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