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_x→0⁺ (cot x - 1/x)

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 Lajapato's rule if necessary. The limit as X approaches 0 from the right of 1 divided by X minus cosicant of X. Awesome. So it appears for this particular problem, we're asked to evaluate this specific limit and that's what we're ultimately solving for. We're also given a hint that we are possibly going to have to use Laha Patal's rule to possibly evaluate this specific limit. So with that said, 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 halves, and D is 2. Awesome. So first off, in order to evaluate the limit, as X approaches 0 from the right, that's what that plus sign and the power means. That means we're evaluating it from the right. So as so in order to evaluate our limit as X approaches 0 from the right of 1 divided by X plus cosicant of X, we must first write cosicant of X as 1 divided by sin of X, and therefore, if we write close can as 1 divided by sin of X, we could rewrite our limit as the limit as X approaches 0 from the right of 1 divided by X minus 1 divided by sin of X. So when we combine like factors, we can go ahead and write that are we can rewrite our limit as the limit. As X approaches 0 from the right of sign of X minus X divided by X multiplied by sin of X. Fantastic. So, moving right along here, we need to note that as X approaches 0 from the right, we need to note that both the numerator, sin of x minus X. And the denominator X multiplied by sin of X will both approach 0, which will result in an indeterminate form. So let's take a moment here to note that an indeterminate form is when we use direct substitution, meaning We pay attention to what the limits being evaluated as. So in this case, X is being so X is approaching 0 from the right. So when we use the direct substitution method, meaning we take this value of 0 and we plug it in for X, so when we plug in this. Value of zero in for X, that will mean we will find our limit when we evaluate it. It's going to be equal to 0 divided by 0, which is an indeterminate form, and also an indeterminate form is infinity divided by infinity. Awesome. And because when we evaluate the limit using direct substitution and we get an indeterminate form, that means we can recall and use Lahapital's rule in order to evaluate and solve for the limit. So to quickly recap here, Once again, so our numerator and our denominator, so if we take sine of 0, minus 0 is going to be equal to 0, and our denominator D for short, so numerators N for short, and denominator is D for short, so 0 multiplied by sin. of 0 is going to be equal to 0. So that gives us an indeterminate form. And because once again there's an indeterminate form as an answer, we need to recall and use Lahapi's rules so we can apply La Hopi's rule to evaluate this limit because of this indeterminate form we get as an answer. So as we should recall, La Hopi's rule states that if the limit results in an indeterminate form, then we can go ahead and write. That the limit as X approaches C, so as the limit as X approaches C of F of X divided by G of X is going to be equal to the limit as X approaches C of F of X divided by G of X, where this prime means it's the first derivative of F of X and then G of X is the first derivative of G of X. And once again, this is provided that the limit on the right side exists, and then this statement will hold true. So with that said, we now need to differentiate our numerator and denominator of our limit. So the derivative of our numerator, which once again is sin of x minus X, is going to be cosine of X minus 1. And the derivative of our denominator is going, so once again, the derivative of our denominator, which is going to be X multiplied by sine of X, is going to be equal to sin of x plus X multiplied by cosine of X. Fantastic. So, now we can apply Lahapa's rules, so we need to take the limit as X approaches 0 from the right of cosine of X minus 1 divided by sine of X plus X multiplied by cosine of X. Fantastic. So now we need to note that as X approaches 0 from the right, that will mean, when we focus once again on our numerator, that will mean that Cosine of 0 minus 1 is going to be equal to 0, and then for our denominator, sine of 0 plus 0, multiplied by cosine of 0 is going to be equal to. 0. So then again, we have an indeterminate form, so that means we have to apply Lajapatal's rule again means essentially we need to keep using Lajapatal's rule until we do not get an indeterminate form as an answer when we evaluate our limit using direct substitution. So with that said, we now need to take the derivative of our numerator and denominator, so we need to take the derivative of Cosine of X minus 1, and when we take the derivative of cosine of X. -1, we will find that the derivative is going to be equal to negatives of X, and the derivative of our numerator, which so we have our numerator, which is going to be negative sin of x, and if we take the derivative of our denominator, which is going to be the derivative of sin of x plus x multiplied by cosine of X, when we perform that derivative. We will find that it's going to be Equal to, so when we start to do the derivative, we'll get, when we start to perform the derivative. We will get cosine of X plus cosine of X plus X multiplied by negatives of X, which will be simply equal to 2 multiplied by cosine of X minus X multiplied by sin of X. Fantastic. So, with that said, we now need to rewrite our limit. So now we have the limit as X approaches 0 from the right. Of negatives of X divided by 2 multiplied by cosine of X minus X multiplied by sin of X. Fantastic. So now, when we use our direct substitution method. We will find that negative sign of 0. is going to be equal to 0. Whereas when we evaluate Our our denominator now, so when we take 2 multiplied by cosine of 0 minus 0 multiplied by sign of 0, we will find that our numerator. It's going to be equal to 0 once again in our denominator, when we evaluate that, it will give us an answer of 2, but since this is not an indeterminant form, we have 0 divided by 2 and 0 divided by 2 means it's going to be equal to 0. So therefore, our limit evaluates to simply 0. And once again this is when we use the direct substitution method. So since x approaches 0 from the right, we plug in a value of 0 in for X. So negatives of 0 is going to be 0 and 2 multiplied by the cosine of 0 minus 0 multiplied by sin of 0 is going to give us an answer of 2. And 0 divided by 2 is 0. So looking at our multiple choice answers, the correct answer, without a doubt 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_x→0⁺ (cot x - 1/x)

Key Concepts

Limits

l'Hôpital's Rule

Cotangent Function

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