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→ 0 (eˣ - sin x - 1) / (x⁴ + 8x³ + 12x²)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to apply Lopita's rule to find the limit of the following function as X approaches 0, and we're given the function H of X is equal to the quantity of 2 multiplied by e to the X minus 2 multiplied by sin of x minus 2 in quantity, divided by the quantity of X to the 5th, plus 2 x cubed plus 3 x squared in quantity. We're given 4 possible choices as our answers. For choice A, we have 1 divided by 3. For choice B, we have 2 divided by 3. For choice C, we have 1 divided by 5, and for choice D, we have 2 divided by 5. Now we're asked to evaluate the limit as X 0 of our function H of X, and we need to apply Lopital's rule to find limit. So we're gonna be looking for the limit. As X approaches 0. Of HF X. And this is going to be equal to the limit. As X approaches 0, of the quantity of 2 multiplied by E to the X minus 2 multiplied by sin of X. Minus 2 in quantity, divided by the quantity of X 5 plus 2 X. Plus 3 x squad in quantity. And so we can try to evaluate this limit by direct substitution, and in doing so, this is going to be equal to 2 multiplied by E 0, minus 2 multiplied by sine of 0, minus 2, and that quantity is divided by the quantity of 0 to 5th power, plus 2 multiplied by 0 cubed. Plus 3 multiplied by 0 squared. And when we evaluate this expression, this is going to be equal to 0 divided by 0. So this here is an indeterminate form. So we're going to apply Lopital's rule to find the limit here, and what we're going to need to do is take a derivative with respect to X of both our numerator and denominator, and then re-evaluate the limit. So this is going to be equal to the limit, as X approaches 0 of the quantity of the derivative with respect to X of the quantity of 2 multiplied by E X minus 2 multiplied by sine of X. Minus 2 in quantity, and that is divided by the derivative with respect to X of the quantity of X 5 plus 2 X cubed. Plus 3 x 2 in quantity. So taking the derivatives, this is going to be equal to the limit. As X approaches 0. Of the quantity, and in the numerator, I want to take the derivative, I will get 2 multiplied by E to the X. Minus 2 multiplied by cosine of X in quantity, and that is divided by the quantity of 5 X to the 4th plus 6 X squad. Plus 6 X. So here we've used the fact that the derivative with respect to X of E to the X is ED X. The derivative with respect to X of sin of X is equal to cosine of X, and we've used the power rule in our denominator. So now, if we try to um evaluate this limit by direct substitution, this is going to be equal to 2 multiplied by e, minus 2 multiplied by cosine of 0, and that quantity is divided by the quantity of 5, multiplied by 0 to the 4th, plus 6 multiplied by 0 squared, plus 6 multiplied by 0. And when we evaluate this expression, we again Have it been equal to 0 divided by 0. Again, it's an indeterminate form. So we're going to again have to take a derivative of both our numerator and denominator, and then re-evaluate the limit. So this is going to be equal to the limit, as X approaches 0. Of the quantity of the derivative with respect to X of the quantity of 2 multiplied by e to the X minus 2 multiplied by cosine of X. In quantity, divided by the derivative with respect to X. Of the quantity of 54. Plus 6 X squared plus 6 X. In quantity. So, if I weighing the derivatives, this is going to be equal to the limit. As X approaches 0. Of the quantity in the numerator, we have 2 multiplied by E to the X. 2 multiplied by minus sign of X in quantity. That is divided by the quantity of 20X cubed. Plus 12 X. Plus 6. So here we've used the fact that the derivative with respect to X of cosine of X is equal to minus sine of X. So, at this point, we can try to evaluate this limit by direct substitution, and in doing so, this is going to be equal to 2 multiplied by E to 0, minus 2 multiplied by minus sine of 0. And that quantity is divided by The quantity of 20 multiplied by 0 cubed. Plus 12. Multiplied by 0. Plus 6. This is going to be equal to. 2 minus 2 multiplied by 0, and that quantity is divided by 0 + 0 + 6. And simplifying this expression, this is going to be equal to 2 divided by 6, which is equal to 1 divided by 3. So this is the answer that we're actually looking for for this problem. And when we compare our answer to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.lim_x→ 0 (eˣ - sin x - 1) / (x⁴ + 8x³ + 12x²)

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

lim_x→ 0 (eˣ - sin x - 1) / (x⁴ + 8x³ + 12x²)