109-112 {Use of Tech} Calculating limits The following limits...

Question

109-112 {Use of Tech} Calculating limits The following limits are the derivatives of a composite function g at a point a.

b. Use the Chain Rule to find each limit. Verify your answer by using a calculator.

$\lim_{h \to 0} \frac{\frac{1}{3 {({(1 + h)}^{5} + 7)}^{10}} - \frac{1}{3 {(8)}^{10}}}{h}$

Accepted Answer

Welcome back, everyone. In this problem, we want to apply Lapita's rule to find the limit of the following function as X approaches 0. G of X equals 1 divided by 2 multiplied by 1 + X 6 + 49 minus 1 divided by 2 multiplied by 59, all divided by X. A says it's 9 divided by 5 to 8th. B 9 divided by 5 to 10th, C-27 divided by 5 to 8th, and D-27 divided by 5 to 10th. Now, before we apply Locuta's rule to find the limit of the function, we first need to establish that we cannot find the limit using direct substitution, OK? We don't want to really waste our time. So let's first check if the direct substitution for our limit, that is when X equals 0, leads to an indeterminate form. Now, by direct substitution, OK. My direct substitution. Then the limit as X approaches 0 of G of X. is going to be equal to 1 divided by 2 multiplied by 1 + 0 raised to the 6 + 4, all raised to the 9th, OK, minus 1 divided by 2 multiplied by 5 to the 9th, all divided by X, which is 0. Now, we've run into a problem here because we can't divide by 0. Anything that's divided by 0 is is going to be undefined, OK? So this is how we know that we'll need to use Lapita's rule to find the limit of the function. Now, what does that rule say? We recall that it basically tells us that if we are finding the limit as x approaches some value c. Of a quotient, in this case, let's call the, the function at the top H of X and the function in the denominator K of X. Then it would be the same thing as finding the limit. As it approaches that value of the derivative of the H of X, the function in the numerator divided by the derivative of K X, the function in the denominator. So let's differentiate our numerator and denominator in our quotient and then find the limit as X approaches 0 for their result. Now Here, OK. Notice that for our denominator, it's probably simpler than our numerator. So we can just write it down beside it here. No the derivative of our denominator, the derivative of X is equal to 1. Next, let's tackle the derivative of the numerator. Notice here that we'll need to use the chain rule. So if we let, OK, let's let our numerator H of X be equal to 1 divided by our expression, which we could rewrite as 2. So we could rewrite here as. Sorry, a half multiplied by 1 + X raise to the 6 + 4 always to the power of -9. OK. Then we can differentiate H of x using the chain rule and basically the chain rule tells us. And first, we're gonna bring down our power, that's -9, OK. Then we are going to multiply it by our expression minus 1 from its power. So that's a half multiplied by 1 + X to the 6 + 4 raise to -9 minus 1, which is -10. And now we're gonna multiply that by the derivative of what's in our inner function, the derivative of 1 plus X-rays to the 6th plus 4. Now if we think about it here. That's going to be equal to 6, OK, multiplied by 1 + X to the 5th, because again, ironically, if we apply the chain rule, we'll bring down the 6, subtract 1 from the power, and multiply the by the derivative of 1 + X which would just be 1. And then the derivative of 4 would just be 0, OK? Now, if we go ahead and simplify this expression, OK. Then we're going to get it to be equal to -27, OK, multiplied by 1 + X raised to the 5th, and all of that is going to be equal, OK, to 1 + X raised to the 6th plus 4, all raised to the point of 10, OK, in our denominator. Know that we have a derivative for the numerator, OK. And we have one for the denominator. We can apply Lapita's rule. Because no, by the rule, that means then that the limit as X approaches 0 for a function G of X. Is going to be the limit as X approaches 0 of the derivative of each term uh the numerator and the denominator. No questions, so that's gonna be this entire function divided by 1, which would just give us back our function. So that's negative 27 multiplied by 1 + X to the 5th, all divided by 1 + X to the 6 + 4 or is to the 10th. Now we can substitute. I don't think we need these brackets here, because now we can substitute our value for our X, OK? So that Let me write it down here. It's gonna be equal to -27 multiplied by 1 + 0 to the 5th. All divided by 1 + 0 to the 6th + 4, to the 10th. No That's -27 multiplied by 1 to the 5th, which is just negative 27, divided by 1 + 0, which is 1 raised to the 6th, which is still 1. So that's gonna be 1 + 4 raised to the 10th and 1 + 4 equals 5. So our answer is -27 divided by 5 to 10th. That's the limit as X approaches 0 of G of X using Lapita's rule. Now if we look back in our answer choices then that tells us D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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