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 - cosⁿ x) / x²

Accepted Answer

Welcome back, everyone. In this problem, we want to find the limit as X approaches 0 of 1 minus the cosine of 3 X 2 divided by X to the 4th. A says it's -3 halves, B 3 halves, C-9 halves, and the D 9 halves. Now, we can first try to find the limit by direct substitution. That's the most straightforward and easiest way, OK? So what that means is that we are going to substitute X equals 0 into our expression. So anywhere we see X, we're going to write 0, OK? So now when we do that. Then it's going to be 1 minus the cosine of 3 multiplied by 0 squared. Divided by 0 to 4th. No, the cores, sorry, 3 multiplied by 0 squared is 0, the cosine of 0 is 1, and 1 minus 1 equals 0. In the denominator, 0 to the 4th is 0, but we've run into a bit of a problem here. No, notice that. Our result is in an indeterminate form because we are dividing by 0. So because we can't divide by 0, that means we can't evaluate our limit by direct substitution. Instead, we need to apply Lapita's rule. What does that rule say? Well, recall that it basically tells us that if we have a limit as X approaches some value of a quotient FX divided by GX. And it's in an indeterminate form, then we can rewrite our limit as x approaches the value of the derivative of the numerator of the quotient divided by the derivative of the denominator. So let's differentiate our numerator and denominator. You apply Lapita's rule and then see if it can help us to find the limit. No, that means we can let F of X or numerator be equal to 1 minus the cosine of 3 X2, OK. And we can let geo x or denominator be equal to X 4th. Then the derivative of our numerator. Let me just clean this up a little bit. The derivative of our numerator is going to be equal to 6 X multiplied by the sign of 3 X squared. And the derivative of our denominator is going to be equal to 4 x cubed. Know that we have those derivatives by applying Lapita's rule, then we can rewrite our limit, as the limit as X approaches 0 of 6 X 3 X squared. Divided by 4 x cubed. Now where do we go from here? Well, notice that 6 and 4 have a common factor of 2, so we could simplify this even more, OK. The limit as x approaches 0 by canceling 2 from our numerator and denominator and also canceling X from our numerator and denominator. And all that's gonna leave us with the limit as X approaches 0 of 3 multiplied by the sign of 3 x 2 divided by 2 X2. Now what do you notice here? Notice that the argument of our sign function and our denominator, they're, they're a little similar. Both of them have X squared in their terms. Is there anything that we can use to relate them in terms of their limits when their argument and their when the argument and the denominator is the same or similar? Well, recall that for the standard limit. If we're finding the limit of Z as it approaches 0, of the sign of Z divided by Z, OK, where Z is the argument of the sine function and the denominator, then it will be equal to 1. So what that means is that if we let. Z be equal to the argument of the sine function 3x2d. If we can rewrite 2 x squared as 3 x 2, then we should be able to apply the standard limit. How are we going to do that? Well, if we take our limit, OK, as X approaches 0 of 3s. 3 x 2 divided by 2 X2, OK. Then if we multiply. Our expression, so the limit as X approaches 0 by 9 halves, OK. Then no, we can write this as, well, we can write this as the sign, OK, of 3 x 2 divided by 3 X2. Because you know, if we think about it here, 9 halves multiplied by, so I think 9/2 multiplied by 1 divided by 3 X squared. In this case, 3 goes into 93 times and now we'll get back 3 in our numerator and 2 X squared in our denominator just like what we had here before. So now, in that case, let me just keep this here for reference. In that case, No, we could rewrite this as 9 houses multiplied by the limit as X approaches 0 of the sign of 3 X squared. Divided by 3 squared. Which we can apply the standard limit to rewrite this then as one. And 9 halves multiplied by 1 equals 9 halves. Thus, this is the final answer, and if we look back at our answer choices, that means D is correct. Thanks for watching everyone. I hope this video helped.

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

lim_x→0 (1 - cosⁿ x) / x²

Key Concepts

Limit of a Function

Taylor Series Expansion

L'Hôpital's Rule

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