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 (x + cos x)¹/ˣ

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 from the right, and we're getting the function G of X is equal to the quantity of 2 X plus cosine of X in quantity, rates to the quantity of 3 divided by X in quantity. We're given 4 possible choices as our answers. For choice A, we have E. for choice B, we have E squared. For choice C, we have E cubed, and for choice D, we have E rated to the 6th. Now, we're asked to find the limit as X approaches 0 from the right of our function G of X using Lopiol's rule. So we need to evaluate the limit as X approaches 0 from the right of G of X. Which is equal to the limit, as X approaches 0 from the right of the quantity of 2 X plus cosine of X. In quantity, raised to the quantity of 3 divided by X in quantity. If we were to evaluate this limit by direct substitution, we would get that this is equal to 1 raise to the power infinity. Since we have 2 X plus cosine of x as our base, and when X is equal to 0, we get the value of 1. And 3 divided by X approaches infinity as X approaches 0 from the right. Now one reason to the power of infinity is an indeterminate form. However, we cannot apply Lopital's rule to this case. So what we're going to do instead is take the natural log of our function G of X and evaluate the limit as X approaches 0 from the right of that function. So we'll have the limit as X approaches 0 from the right of the natural log of G of X. Which is equal to the limit, as X approaches 0 from the right of the natural log of the quantity of 2 X plus cosine of X in quantity, raised to the quantity of 3 divided by X. Now, we can rewrite this expression using our properties of logarithms, so we're going to use the power rule for logarithms. So this is going to equal to the limit, as X approaches 0 from the right of 3 divided by X, multiplied by the natural log of the quantity of 2 X plus cosine of X. In quantity. Now, if we were to try to evaluate this limit by direct substitution, then this was going to be equal to the 3 multiplied by the natural log of 1. Divided by 0. Which is equal to 0 divided by 0. So here we still have an indeterminant form, but we can actually apply Lopital's rule to this expression. So we call for Lopital's rule, we'll need to take the derivative with respect to X of both our numerator and denominator of our fraction here, and then re-evaluate the limit. So this is going to be equal to the limit as x approaches 0 from the right. Of the derivative with respect to X, of the quantity of 3 multiplied by the natural log of the quantity of 2 X plus cosine of X in quantity. And that is divided by the derivative with respect to X of X. So, evaluating the derivatives, this is going to be equal to the limit. As X approaches 0 from the right. And in the numerator, we'll have the derivative with respect to X of 3 multiplied by the natural log of the quantity of 2 X plus cosine of X. So since 3 is constant, it can move through the derivative, and when we take the derivative of the natural log function, we get 1 divided by its argument. So we're gonna have 3, multiplying the quantity of 1 divided by the quantity of 2 X plus cosine of X. However, we need to multiply that fraction by the derivative of the arguments. We need to apply our chain rule. And so we're going to need to multiply this by the derivative with respect to X of the quantity of 2 X plus cosine of X, which when we take that derivative, we get 2 minus sine of X. In quantity, and that is divided by the derivative of X with respect to X, which is 1. So simplifying this expression, this is going to be equal to the limit, as X approaches 0 from the right. Of 3, multiplying the quantity of 2 minus sine of X. In quantity, divided by the quantity of 2 X plus cosine of X. So now we can evaluate this limit by direct substitution, so this is going to be equal to. 3, multiplied the quantity of 2 minus sign of 0 in quantity, divided by the quantity of 2 multiplied by 0, plus cosine of 0. So simplifying this expression, this is going to be equal to 3 multiplied by the quantity of 2 minus 0 in quantity, divided by 0 + 1. And when we evaluate this expression, this is just going to be equal to 6. So now what we need to do is relate the limit that we just calculated to our original limit. So, coming back to our original limit, which was the limit. As X approaches 0 from the right of the quantity of 2 X. Plus cosine of X in quantity. Raised to the quantity of 3 divided by X in quantity. And we can rewrite this using exponentials, so this is going to be equal to the limit, as X approaches 0 from the right. Of the quantity of e raised to the quantity of the natural log of the quantity of 2 X plus cosine of X in quantity raised to the quantity of 3 divided by X in quantity. Now we just found that the limit as X approaches 0 from the right of the natural log of the quantity of 2 X plus cosine of X in quantity raised to the quantity of 3 divided by X is equal to 6. So this here is going to be equal to. He raises to the 6th power, and that is actually the answer that we're looking for. So now that we have our final answer, we can compare it to our answer choices, and we see that the correct answer choice corresponds to answer choice D. 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 (x + cos x)¹/ˣ

Key Concepts

Limits

L'Hôpital's Rule

Exponential Functions

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