17–83. Limits Evaluate the following limits. Use l’Hôpital’s R...

Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_Θ→π/2⁻ (tan Θ)ᶜᵒˢ ᶿ

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says to apply Lipita's rule to find the limit of the following function as X approaches 0 from the right, and we're given the function F of X is equal to the quantity of cotangent of X in quantity raised to the sine 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 2. Now we're asked to find the limit as X approaches 0 from the right of our function F of X using Lopital's rule. So we need to find the limit. As X approaches 0 from the right of F of X. Which is equal to the limit, as X approaches 0 from the right. Of the quantity of cotangent of X. In quantity raised to the sign of X. Now if we were to try to evaluate this limit by direct substitution, this would be equal to infinity rates to the power of 0. Since cotang of X approaches infinity as X approaches 0 for the right, and the sign of 0 is 0. So here we have an indeterminate form. However, we cannot apply Lopital's rule to something that has this indeterminate form. So, what we're going to do is rearrange our equation here to get a form that we can actually apply Lopita's rule. So we're going to do is take the natural log of our function of FF X and then find the limit of that quantity. So we'll have the limit. As X approaches 0 from the right of the natural log of F of X. Which is equal to the limit, as X approaches 0 from the right. Of the natural log of the quantity of cotangent of X in quantity raised to the sine of X. So, we can rewrite this expression using our properties of logarithms. So here we'll use our power rule for logarithms. This is going to be equal to the limit. As X approaches 0 from the right of sign of X. Multiplied by the natural log of cotangent of X. Now, ultimately we want to apply Lopita's rule, and in order to apply Lopita's rule, we're going to need a fraction. So we're going to rewrite the sine of X in terms of co-sequence. So this is going to be equal to the limit. As X approaches 0 from the right of the natural log of cotangent of X. Divided by Cosequent of X. So recall that sin of X is equal to 1 divided by cosequent of X. And so now, if we try to evaluate this limit by direct substitution, this is going to be equal to infinity divided by infinity. Since both cotangent of X and cosse of X approach infinity as X approaches 0 from the right. So here we still have this indetermined the form, but we can actually apply Lopita's rule here in this case, since we now have a fraction. So we call to apply Lopita's rule, we need to take the derivative with respect to X of both our nitter and denominator, and then re-evaluate the limit. So this is going to be equal to the limit. As X approaches 0 from the right of the derivative with respect to X of the natural log of cotangent of X. That is divided by the derivative. With respect to X, Of Cosequant of X. So, at this point we just need to take the derivatives. So this is going to be equal to the limit. As X approaches 0 from the right. And in the numerator, we'll have the derivative with respect to X of the natural log of cotangent of X. So recall that when you take the derivative of the natural log function, you get 1 divided by its argument. So we'll have 1 divided by cotangent of X. But we'll need to multiply that quantity by the derivative of the argument, since we need to apply our chain rule, so we'll have the derivative with respect to X of cotangent of X, which is equal to minus cosequent squared of X. And then this is divided by. The derivative with respect to X of cosequent X, which is equal to cosequent of X sorry, minus cosequent of X. Multiplied by cotangent of X. So now we just need to simplify this expression, so this is going to be equal to the limit. As X approaches 0 from the right. Of Cosequant of X. Divided by Cotangent where of X. Now, we want to rewrite everything in terms of sine and cosine, and so recall that cosequence of X is equal to 1 divided by sine of X, and cotangent of X is equal to cosine of X divided by sin of X. So some find this expression, this is going to be equal to the limit. As X approaches 0 from the right of sign of X. Divided by cosine squared of X. So now we can evaluate the the limit by direct substitution. And so doing so, this is going to be equal to the sign of 0, divided by. Cosine squared of 0. And since the sin of 0 is 0, and the cosine of 0 is 1, this is going to be equal to 0 divided by 1 squared, which is equal to 0. So now what we need to do is relate the limit that we have found here back to our original limit. So recall that we are looking for the limit. As X approaches 0 from the right of the quantity of cotangent of X. In quantity raised to the sign of X. And we can rewrite this using exponentials. So this is going to be equal to the limit. As X approaches 0 from the right, of E raised to the natural log of the quantity of cotangent of X. In quantity raised to the sign of X. Now we just found that the limit as x approaches 0 from the right of the natural log of the quantity of cotangent of X in quantity rates to the sin of X is equal to 0. That means that this here is going to be equal to E rates to the 0, which is equal to 1. And so this is the answer that we're looking for for this problem. And so when we compare our answer here 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.

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_Θ→π/2⁻ (tan Θ)ᶜᵒˢ ᶿ

Key Concepts

Limits

L'Hôpital's Rule

Trigonometric Functions

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