Use l’Hôpital’s Rule to find the limits in Exercises 85–108.

Question

102. lim(x→0) (x sin(x²))/(tan³x)

Accepted Answer

Hello there, today we are 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. Find the limit using La Hopatal's rule. The limit as x approaches 0 of x multiplied by sin of x divided by tangent squared of x. OK. So it appears for this particular prom, we're asked to take the limit that is provided to us by the prom itself, and we're asked to find the limit using Lahopatal's rule. So now that we know that we are ultimately trying to find what this limit is using Lahopatal's rule, our first step that we need to take is we need to check if there is an indeterminate form. So how do we do that? So, first off, let us write down our limit. So we have the limit as X approaches 0 of X multiplied by sin of X. Divided by tangent squared of x. And this will be equal to, so we're told that the limit as x approaches 0. So let us take the 0 and plug in a value of 0 in for X. So we will find that it's equal to square brackets of. 0. So once again, we're plugging in a value of 0 infra x, so 0 multiplied by sin of 0. So 0 multiplied by sin of 0, divided by Tangent square of 0, so tangent. Square, so tangent square of 0. And when we evaluate this expression, we will find that it'll be equal to 0 divided by 0 in square brackets, and 0 divided by 0 is indeed an indeterminate form, and because we have an indeterminate form, we must apply La Hopatal's rule. So, as we should recall. In order to apply Lahapatal's rule, we need to recall that we need to differentiate the numerator and denominator separately with respect to X. And after applying Lahapatal's rule, if the expression still results in an indeterminate form, we need to repeat the differentiation process until a determinate form is achieved. So sometimes you have to Apply Lahapatal's rule 2 to 3 different times in order for it to achieve a determinate form, because you have to keep applying Lahapatal's rule until a determinate form is achieved. So it can't be 0 divided by 0 or infinity divided by infinity. So with that said. Let us apply Lahapatal's rule, so we need to take the limit as X approaches 0 of. D by Dx of x multiplied by sin of x, all divided by d by dx of tangent squared of x, which is equal to the limit as x approaches 0 of x multiplied by. Cosine of x plus sin of x, all divided by 2 multiplied by tangent of x multiplied by c square of x, and this will be equal to square brackets of, so now we're going to plug in a value of 0 in for x will be equal to 0 multiplied by cosine of 0 plus. Sign of 0. All divided by 2 multiplied by. Tangent. Of 0, so tangent of 0, multiplied by, so it's gonna be multiplied by C can't, so it's gonna be multiplied by C can't square of the, so it's gonna be gonna be equal to c square of 0. Fantastic. And this will be equal to, when we simplify, 0 divided by 0, which is unfortunately an indeterminate form. So we need to apply La Hopatal's rule again. And when we do that, we will find that we could write that the limit as X approaches 0 of D by DXX multiplied by cosine of X plus sin of X, all divided by D by DX 2 multiplied by tangent of X multiplied by C2 X. Which will be equal to the limit as x approaches 0 of. Negative x multiplied by sin of x plus cosine of x plus cosine of x, all divided by 2 multiplied by the parentheses of tangent of x multiplied by the parentheses of 2 multiplied by c can't. Of x multiplied by c x multiplied by tangent of x + plus c 2x multiplied by c square. Of X, and don't forget your final parenthesis. Fantastic. And this. When we simplify, will be equal to the limit as x approaches 0x multiplied by sin of x plus 2 multiplied by cosine of x, all divided by 2 multiplied by parentheses of 2 multiplied by tangent square of x multiplied by c 2 x plus. C can't to the power of 4 X. Fantastic. So now we can plug in a value of 0 in for X, and when we do, we will find that it's equal to square brackets of 0 multiplied by sin of 0, so it's gonna be multiplied by sin. Of 0, so sign of 0. And we need to add 2 multiplied by Osin of 0. All divided by 2 multiplied by parentheses of 2 multiplied by tangent 20 multiplied by c2 of 0. Plus C can't to the power of 4 of 0. Fantastic. So This will be equal to square brackets of, when we begin to simplify, 0 + 2 multiplied by 1, all divided by 2, multiplied by parentheses of 2 multiplied by 0, multiplied by 1 plus 1, and don't forget your parentheses, which will be equal to square brackets of 2 divided by 2 multiplied by 1. Which will mean that our final answer. Without a doubt will be multiple choice, well. There is no multiple choice answer, it'll just be simply one. There we go, so our final answer is one. Hooray, we did it. So. Going back to the top to recap what we've learned, we could safely say that the evaluated limit, our final answer for the limit is simply one. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Use l’Hôpital’s Rule to find the limits in Exercises 85–108.102. lim(x→0) (x sin(x²))/(tan³x)

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Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
102. lim(x→0) (x sin(x²))/(tan³x)