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→π / 2- (sin x) ^tan x

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the limit, use Lopital's rule if necessary, and we're given the limit as X approaches 0 from the left of cosine of X-rays to the cotangent of X. We're given 4 possible choices as our answers. For choice A, we have 0, for choice B, we have 1. For choice C, we have minus 1, and for choice D, we have D and E, which stands for does not exist. Now, we need to just evaluate the limit. As X approaches 0 from the left, of cosine of X. Raised to the cotangent of X. Now, if we try to evaluate this limit by direct substitution, this is going to be equal to. 1, raised to the minus infinity, since cosine of zero approaches 1, and cotangent of 0 approaches minus infinity as we approach 0 from the left, which we can rewrite this expression as being equal to 1 raise to the infinity. Which is an indeterminate form. So, what we're going to do is actually calculate a slightly different limit. We're going to take the natural log of this function here and calculate its limit as X approaches 0 from the left. So, we'll have the limit. As X approaches 0 from the left of the natural log of cosine of X. Raise to the cotangent of X. And we can rewrite this using our properties of logarithms as being equal to the limit, as X approaches 0 from the left. Of cotangent of X, multiplied by the natural log of cosine of X. And we can rewrite cotangent of X as one divided by tangent of X, but this is going to be equal to the limit. As x approaches 0 from the left of the natural log of cosine of X. Divided by tangent of X. Now, at this point, if we try to evaluate this limit by direct substitution, we will get the cosine of 0, which is 1, and then the natural log of 1 is 0, so this is going to be equal to 0, divided by the tangent of 0, which is 0, which gives us 0 divided by 0, which is again an indeterminate form. But here at this point, we can actually apply Lopital's rule. So, we call for Lopital's rule that I need to take a derivative with respect to X of both the numerator and denominator, and then re-evaluate the limit. So, doing that, this is going to be equal to the limit. As X approaches 0 from the left. Of the derivative with respect to X. Of the quantity of the natural log of cosine of X. In quantity Divided by the derivative with respect to X. Of the quantity of tangent of X in quantity. So, evaluating those derivatives. This is going to be equal to the limit. As X Approaches 0 from the left. Of and in the numerator, we'll have the derivative with respect to X of the natural log of cosine of X. So recall when you take a derivative of the natural log function, you get 1 divided by the argument. So I'll have 1 divided by cosine of X. But I'll need to multiply that by the derivative of our argument, so basically applying our chain rule, and the derivative of cosine of X is equal to minus sine of X. And that is divided by the derivative with respect to X of tension of X, which is de squared of X. So at this point, we can uh try to use direct substitution to evaluate this limit. In doing so, this is going to be equal to 1 divided by cosine of 0. And that quantity is multiplied by the quantity of minus sign. Of 0. And that is divided by. Sequent squared of 0. Now, cosine of 0 and sequent of 0 are both equal to 1, and the sign of 0 is equal to 0. So this expression evaluates to being equal to 0. So now we need to relate the limit that we just found to our original limit. So the limit As X approaches 0 from the left. Of cosine of X raised to the cotangent of X. is equal to The limit as X approaches 0 for the left, of E. Raised uh to the quantity of the natural log. Of the quantity of cosine of X. Raised to the cotangent of X. So when we evaluate this limit, we already found the limit as X approaches 0 from the left of the natural log of cosine of X, raised to the cotangent of X, so this is going to be equal to E to 0, which is going to be equal to 1. So this here is the answer that we're looking for. And We can, when we compare this to our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

60–81. Limits Evaluate the following limits. Use l’Hôpital’s Rule when needed. lim_x→π / 2- (sin x) ^tan x

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60–81. Limits Evaluate the following limits. Use l’Hôpital’s Rule when needed.
lim_x→π / 2- (sin x) ^tan x