Evaluate the following limits. Use l’Hôpital’s Rule when it is...

Question

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

lim_Θ→π/2 (tan Θ - secΘ)

Accepted Answer

Below there today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the limit. Use Lahapatal's rule if necessary. The limit as theta approaches 0 from the right of cotangent of theta minus cosicant of theta. Awesome. So it appears for this particular problem we're trying to evaluate this particular limit, and we're given a hint that we're to use Lajapathal's rule if necessary in order to perform this evaluation of this particular limit. So with that said, now that we know that we're trying to evaluate the limit, let's read off our multiple choice answers to see what our final answer might be. A is -12, B is 0. C is 3 divided by 2. And D is 5 divided by 2. Awesome. So, first off, as we should recall a note, in order to evaluate the limit, and we're trying to evaluate the limit as theta approaches 0 from the right of cotangenta theta minus cosicana theta, we must first we need to first understand the behavior of cotangenta theta and Cosicana Theta as theta approaches 0. And i from the right. So we need to figure out the behavior of cotangenta theta and Cosicana theta. So let us note that cotangent of theta is equal to cosine theta divided by sin of theta. And let us also note and recall that. Can of theta is equal to 1 divided by sign of theta, as we should recall from trig identities. So as theta approaches 0 from the right, so we can write this as theta, as theta approaches 0 from the right. Sine of theta approaches 0 from the right, and, cosine of theta. So we could say that and cosine theta approaches 1. So that will mean that cotangent of theta is equal to cosine theta divided by sin theta is equal to -1 divided by 0 +, which is equal to infinity. And we can also say that cosicant a theta is equal to 1 divided by sin of theta. is equal to 1 divided by 0 to the power of plus 0 from the right. So this 0 to the power of plus, so this plus sign in the power means they're approaching 0 from the right. Awesome. And then this will be also equal to Infinity. So now we need to combine like terms. So we need to note that the expression becomes cotangent of theta minus cosicant of theta is going to, so we're gonna say cotangent. Of theta minus cosicant of theta is going to approach infinity minus infinity. Which will be an indeterminate form, and because it's in an indeterminate form, we need to recall and use Lajapaal's rule. So as we should recall and use Lahapa's rule, we need to, in order to do, or rather, in order to use Lajapatel's rule, we need to rewrite the expression in order to apply Lahapatal's rule. So therefore, we can go ahead and write that cotangent of theta minus cosicant of theta. is equal to cosine theta divided by sin of theta minus 1 divided by sin of theta, which is going to be all equal to cosine theta minus 1 divided by sin of theta when we do a little bit of simplifying. So now we need to note that as theta approaches 0 from the right, both the numerator and denominator will approach 0. So therefore, as a result, we can go ahead and write that cosine of theta minus 1 will go to 1 minus 1, which equals 0, and Sign of theta will approach 0, so cosine theta -1 will approach 1 minus 1, which is 0, and sine theta will approach 0. So thus this once again, which gives us a 0 divided by 0 indeterminate form, which means we have to apply Lajapato's rule again by differentiating the numerator and denominator. So, Let's do the let's take the derivative of the numerator. So when you take D by D theta, which means we're differentiating with respect to theta of cosine theta minus 1. So once again, this is the numerator, and when we take the derivative of the numerator, we'll get negative sine of theta. And if we take the derivative of the denominator, which is D by D theta of sin theta, we will find that it's equal to cosine of theta. So now we can evaluate our new limit, so we need to recall and note that as the as theta. So you go ahead and write that, you could say that as theta approaches 0 from the right. And we can say that as sign of theta approaches 0, and that as cosine of theta approaches 1, the limit will become as a result, the limit as X approaches, or rather, it's gonna be the limit as theta. As so we're gonna say the limit as theta approaches 0 from the right, we can go ahead, let this say that the limit as theta approaches 0 from the right of negative sign of theta, divided by cosine of theta. It's going to be equal to when you plug in a value, so this is very, very important. So since we note that the limit states that theta is approaching 0 from the right, we just need to plug in a value of 0 in for theta, and when we do that, we will find that it's equal to 0 divided by 1, which will give us an answer of 0 when we plug that into our calculator, and that is our final answer. That's it. We did it. We solved for this problem. So therefore, without a doubt, our final answer has to be multiple choice answer letter B, which once again is 0. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.
lim_Θ→π/2 (tan Θ - secΘ)

Key Concepts

Limits

l'Hôpital's Rule

Trigonometric Functions

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