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_x→∞ x³ (1/x - sin 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. Apply Lahapatal's rule to find the limit of the following function as X approaches infinity. F of X is equal to 4 x 3 multiplied by pi divided by 2X minus sign of pi divided by 2 X. Fantastic. So it appears we're asked to apply Lahapa's rule to this limit of for the function F of X. We're trying to find the limit of this function F of X using Lajapa's rule as it approaches. Affinity, so as X approaches infinity. So essentially, once again, we're trying to use Lahapatal's rule to evaluate this limit of F of X as X approaches infinity. That is what we're ultimately trying to do. So with that said, our limit that we're trying to evaluate is the limit as X approaches infinity of or X to the power of 3 multiplied by pi divided by 2 X minus sign of I divided by 2 X. Uh, yes, awesome. So now, Our next step is we need to recall and recognize the fact that the direct substitution of X equals infinity will lead to an indeterminate form which, as we should recall, let's make a note off to the side here that an indeterminate form means we receive an answer when we use direct substitution method of 0 divided by 0 or infinity divided by infinity. So an indeterminate form is either 0 divided by 0 or infinity divided by infinity. Once again, when you use the direct substitution method, and we're told that x approaches infinity, so we're using when we use direct substitution methods, since we're told once again that x approaches infinity, we're plugging in a value of infinity in for x straight into our equation. Inside of our limit, the expression for a limit, and when we plug in a value of infinity in for X, that means we will get an indeterminate form as an answer, which because it's an indeterminate form as an answer, that means we can simply use Lahapatal's rule. To solve for that answer, because you can only use La Hapaal's rule if when you do the direct substitution method and it leads to an indeterminate form as an answer. So, with that said, so when we use the direct substitution once again of X equals infinity, it gives us an indeterminate form, so we use Lahapital's rule. So with that said, we could go ahead and write that the limit as X approaches infinity. Of 4 X to the power of 3 multiplied by I divided by 2 X minus sine of pi divided by 2 X, and this is going to be equal to Or multiplied by infinity to the 3 minus sign of pi divided by 2 multiplied by infinity. Once again, this is just to reiterate our point when we plug in a value of infinity in for X. And when we do that, we will find that it's equal to infinity, multiplied by 0 minus sign of 0. Which will give us an answer of infinity multiplied by 0. So Lahavital's rule is not directly applicable to the indeterminate form, infinity multiplied by 0. However, this form can be written as 0 divided by 0 or infinity divided by 0. So in order to do that, we need to say that Y is equal to pi divided by 2 X, which means that X is equal to pi divided by 2 Y. So then we could go ahead and write that F of pi divided by 2 Y is equal to 4 multiplied by divided by 2 to the power of 3. And then we need to multiply this by Y minus sign of Y. Fantastic. So now we need to go ahead and write that. F of pi divided by 2 Y is equal to 4 multiplied by the power of 3, divided by 8 of 3, multiplied by Y minus sign of Y. Fantastic. So now, We can simplify this expression, and we can go ahead and write that F of pi divided by 2, Y is equal to divided, so it's gonna be pi to the power of 3 divided by 2, and then we need to take pi to the power of 3 divided by 2, and we need to multiply it by. Y minus sign of Y. Divided by Y to the power of 3. And now at this stage we must recall a note that as X approaches infinity, we need to note that Y will approach 0 from the right. So this positive sign in the power means that we're moving, so we're as Y approaches 0 from the right. Fantastic. So now, With that said, we could go ahead and write that as the limit as Y approaches 0 from the right of pi to the power of 3 divided by 2 multiplied by Y minus sine of Y. Divided by the power of 3 is going to be equal to 3 divided by 2 multiplied by the limit as Y approaches 0 from the right of Y minus sign of Y. Awesome divided by Y to the power of 3. Fantastic, we are on a roll. So now, with that said, we now need to evaluate the limit as Y approaches 0 from the right, so as Y approaches 0 from the right, we can go ahead and write that. I to the power of 3 divided by 2 multiplied by the limit as Y approaches 0 from the right of Y minus sign of Y. Divided by the power of 3 is going to be equal to 13 divided by 2 multiplied by 0 minus sign of 0, and note once again that we're plugging it, we're using direct substitution, so we're plugging in a value of 0 in for Y. Divided by 0 to the power of 3. So this means when we evaluate this, we will get an answer of 0 divided by 0. As you can see. Which is awesome because now we can use Lajapatal's rule officially. So in order to apply Lajapatal's rule, as we should recall, we need to differentiate the numerator and denominator separately with respect to Y. And after applying Lajapatal's rule, we need to figure out whether or not the expression still results in an indeterminate form, and if it does, we need to keep repeating this differentiation process. Until a determinant form is achieved. So we need to keep doing this differentiation process until we do not get an answer of zero divided by 0 or infinity divided by infinity. So with that said, we need to recall that D by DY of sign of Y is going to be equal to cosine of Y, fantastic, and we need to recall and use the power rule, which, as we should recall D by D Y. Of the power of N is equal to N multiplied by Y the power of N minus 1. So with that in mind, we need to go ahead and write. That I divided by 2 multiplied by the limit as Y approaches 0 from the right. Of D by D Y, so when you take D by Dy of Y minus sign of Y. And this is gonna be all divided by D by D Y. Of the power of 3, which is gonna be equal to pi to the power and don't forget. That it's pi to the power of 3, so don't forget your power of 3. So it's pi to 3 divided by 2 multiplied by the limit, as Y approaches 0 from the right of D, D by Dy of Y minus sign of Y divided by D by Dy of Y to the power of 3, which is equal to 3. Divided by 2, multiplied by the limit as Y approaches 0 from the right of Y to the power of 1 minus 1 minus cosine. Of Y divided by 3, multiplied by Y to the power 3 minus 1. And this is going to be all equal to pi to the power of 3. So you need to take this to say that is equal to pi 3 divided by 6, multiplied by the limit as Y approaches 0 from the right. Of 1 minus cosine of Y divided by Y to the power of 2. So now we need to evaluate the limit once again as Y approaches 0 from the right. So let me do that, we will take that pi. To the power of 3 so when you take. I to the 3 divided by 6 multiplied by the limit as Y approaches 0 from the right of 1 minus cosine of Y divided by Y 2, which is going to be equal to 13 divided by 6 multiplied by 1 minus cosine of 0. These are plugging in a value of 0 in for Y divided by 0 to the power of 2. Fantastic, which unfortunately when we plug this into our calculator or you could just look at it and recognize the fact that yet again we get an indeterminant form, we get 0 divided by 0 as an answer. So we have to keep different we have to differentiate one more time. So once again, since the expression still results in an indeterminate form, we need to apply Lahapa's rule again by differentiating both the numerator and the denominator. We need to recall that D by Dy of the cosine of Y is going to be equal to negatives of y, and we also need to recall and use the constant rule which states that D by Dy of some value C constant value C is equal to 0. That's what C denotes as some constant value. So with that said, we can go ahead and write without a doubt that I the power of 3 divided by 6. Multiplied by the limit as Y approaches 0 from the right. Of D by Dy of 1 minus cosine of Y. Divided by D by Dy of Y to the power of 2. It's going to be equal to pi to the power of 3 divided by 6 multiplied by the limit as Y approaches 0 from the right. Of 0 minus negative sign of Y. Fantastic, divided by 2 multiplied by Y2 minus 1, or you could just say 2, so 2 to the power of, so we could say 2 to the power of 2 minus 1. So this will be all equal to the power of 3 divided by 12, multiplied by the limit as Y approaches 0 from the right of sign of Y divided by Y. Fantastic. But, as you can see when we evaluate the limit as Y approaches 0 from the right, so that means when we once again plug in a value of 0 in for Y, we will find that pi to the power of 3 divided by 12 multiplied by Sign of 0 divided by 0 just gives us an answer of 0 divided by 0, which once again is an indeterminate form, which we do not want that. So we have to differentiate one more time. So, Essentially, we need to we now need to differentiate one last time, hopefully. So we need to take pi to the power of 3 divided by 12, multiplied by the limit as Y approaches 0 from the right of D by Dy of sign of Y, so we need to take parenthesis, sign of Y. Divided by D by Dy of Y. So when we do that, we will find that it's equal to 1 to the power of 3 divided by 12 multiplied by the limit as Y approaches 0 from the right of cosine of Y, because once again when you perform the differentiation, divided by Y to the power of 1 minus 1. Which will mean that this is gonna be all equal to I to the power of 3 divided by 12 multiplied by the limit as Y approaches 0 from the right of cosine of Y divided by 1, which is going to be equal to the power of 3, divided by 12 multiplied by the limit as Y approaches 0 from the right. Of cosine of y. So when we evaluate this expression, when we plug in a value of 0 in for Y, we will find that we will get an answer of pi to the power of 3 divided by 12 multiplied by cosine of 0, which cosine of 0 is 1, so that means our final answer, without a doubt has to be pi to the power of 3 divided by 12. And that's it. That is our final answer. Hooray, we did it. So, without a doubt, looking at the problem itself, we will determine that our final answer, when we evaluate our limit, will have to be equal to drumroll, ie to the power of 3 divided by 12, and that is our evaluated limit. Hooray, we did it. So one more time, so i to the power of 3 divided by 12 is our final answer for this evaluated limit. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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

lim_x→∞ x³ (1/x - sin 1/x)

Key Concepts

Limits

L'Hôpital's Rule

Asymptotic Behavior

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