Indeterminate Powers and Products
Find the limits in Exe...

Question

Indeterminate Powers and Products

Find the limits in Exercises 53–68.

66. lim (x → 0⁺) x (ln x)²

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. Evaluate the limit. The limit evaluated at X approaches 0 from the right of X multiplied by the parenthesis of the natural log of X to the power of 3. OK, so it appears for this particular problem we're ultimately asked to evaluate the limit that is given to us by the prom itself. So now that we know what we're ultimately trying to do or what we're ultimately trying to solve for, let's take a moment to read off our multiple choice answers to see what our final answer may be. A is 0, B is 1, C is E to the power of 3, and D is 3E. OK. So our first step that we need to take is we need to let capital L equal the limit evaluated from or the limit as X approaches 0 from the right. That's what that positive, that plus sign and the power means. It means that we're approaching 0 from the right. If it was a minus symbol, then that would mean it'd be approaching 0 from the left. So capital L equals the limit, as X approaches 0 from the right of X multiplied by the parenthesis of the natural log of X to the power of 3. Fantastic. Our second step now is we need to evaluate the limit, so we need to take capital L as equal to square brackets of approaching 0 from the right, multiplied by parentheses the natural log of 0, specifically approaching 0 from the right. To the power of 3 in and then don't forget your square bracket on the end, which will be equal to the square bracket of 0 multiplied by negative infinity to the power of 3, which will be equal to square bracket of 0 multiplied by parenthesis negative infinity. Fantastic, which this will give us an indeterminate answer. And because we received an indeterminate answer, this will require us to use algebraic manipulation. So we will need to state that capital L is equal to the limit, as X approaches 0 from the right of the natural log of X to the power 3 divided by 1 divided by X. Which will be equal to square brackets of parentheses, the natural log of parentheses approaching 0 from the right. All to the power of 3 divided by 1 divided by 0 or approaching 0 from the right, and don't forget your square bracket on the end, which will be equal to square brackets of negative infinity divided by positive infinity. So negative infinity divided by positive infinity yet again gives us an indeterminate answer and because once again we received another indeterminate answer, we need to recall and apply Lajapathal's rule. So as we should recall, in order to apply the Lahapatal's rule, we need to differentiate the numerator and denominator separately with respect to X. And after we apply Lajapathal's rule, we need to note that if the expression still results in an indeterminate form, we need to repeat the differentiation process until a determinant form is achieved as an answer. So with that said. We need to go ahead and state that capital L is equal to the limit, as X approaches 0 from the right of D by DX of parenthesis the natural log of X to the power of 3 divided by D by DX 1 divided by X. Which will be equal to the limit as x approaches 0 from the right of 3 multiplied by parentheses the natural log of x 2 multiplied by 1 divided by X in parentheses, all divided by negative 1 divided by X to the power of 2. Fantastic. So following along here, we need to note that this will be equal to the limit as X approaches 0 from the right of 3 multiplied by parenthesis, the natural log of X to the power of 2 divided by negative 1 divided by X. Which this will be equal to square brackets of 3 multiplied by parentheses, the natural log of approaching 0 from the right. Alter the power of 2 divided by negative, 1 divided by approaching 0 from the right. Which When we plug this in, so once again, when we start to solve, we will, and, you know, since we plugged in our value for approaching 0 from the right, we will find that this will be equal to square brackets of drum roll, 3 multiplied by negative infinity to the power 2 divided by negative infinity. Which will ultimately give us an answer of that's equal to square brackets of positive infinity divided by negative infinity which once again we are given we are rather we find an indeterminate answer so we need to apply La Hapital's rule again. So we need to state that capital L is equal to the limit, as X approaches 0 from the right of D by D X 3 multiplied by parenthesis of the natural log of X to the power of 2, all divided by D by DX of negative 1 divided by X. Fantastic. So, similarly, we're going to follow the same methodology, but I'm going to skip a few steps because we should know how to differentiate by this stage. So, following our differentiation scales, we will find that it will be equal to when we differentiate the limit. As X approaches 0 from the right of 6 multiplied by the natural log of X divided by 1 divided by X. So now at this stage, we need to plug in our value of approaching 0 from the right into our expression. So we will find that this is equal to square brackets of 6 multiplied by the natural log of approaching 0 from the right. Divided by 1, divided by approaching 0 from the right. Which when we evaluate this, we will find that it's equal to square brackets of negative infinity divided by positive infinity, which once again gives us an indeterminate answer. So once again we have to apply La Hapital's rule. So when we do that, we will find that capital L is equal to the limit, as X approaches 0 from the right. Of D by D X of 6 multiplied by the natural log of X divided by D by DX of 1 divided by X, which will be equal to when we perform our differentiation, we will find that it'll be equal to the limit as X approaches 0 from the right of -6 X. Which will mean that L will be equal to square brackets of -6 multiplied by 0 so approaching 0 from the right, since we're plugging in our value, since our limit is as X approaches 0 from the right, so we have to plug that in for our value of X. So it'll be -6 multiplied as, you know, 0, approaching 0 from the right. Which we will find when we evaluate this expression, we will find that it's simply equal to 0 because -6 multiplied by 0 is 0. So therefore, without a doubt, we can go ahead and write that capital L is equal to the limit as X approaches 0 from the right. Of X multiplied by the natural log of X in parenthesis to the power of 3 will be simply equal to 0. Hooray, we did it. So looking at our multiple choice answers, our final answer without a doubt will be multiple choice answer, letter A, which is 0. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Indeterminate Powers and ProductsFind the limits in Exercises 53–68.66. lim (x → 0⁺) x (ln x)²

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Indeterminate Powers and Products
Find the limits in Exercises 53–68.
66. lim (x → 0⁺) x (ln x)²