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

Question

Indeterminate Powers and Products

Find the limits in Exercises 53–68.

60. lim (x → 0) (e^x + x)^(1/x)

Accepted Answer

Welcome back, everyone. Compute the limit. Limit as X approaches 0 of E to the power of 2 X plus X, all erased to the power of 1 divided by X. A says 1 divided by E cubed, B 1 divided by Esquared, CE2, and D E cubed. For this problem, let's begin by rewriting the limit. Limit as X approaches 0 of E to the power of 2 X plus X rates to the power of 1 divided by X. Let's begin the problem by substituting X equals 0. So we get E to the power of 0 plus 0. Raise the power of 1 divided by 0. We're using brackets because technically, this is not correct, right? Division by zero is undefined. But overall, we want to understand what type of indeterminate form we get. So we get a 1, a race to the power of 1 divided by 0, which is infinity. So one divided by one race of the power of infinity is an indeterminate form. And one of the ways to solve for such a limit is to take the natural logarithm of the limit. If we say that L is equal to the original limit, we can take LN of L to apply the properties of logarithms and evaluate the limit. So the natural log of the limit is going to be limit as x approaches 0 off LN of E to the power of 2 X plus X, raises the power of 1 divided by X, which allows us to bring down the exponent, right? We're going to apply the power rule of logarithms. This is going to be equal to limit as x approaches 0 of. LN of E to the power of 2 x plus x divided by X. And now once again, we're going to attempt direct substitution, which gives us a line of E to the power of 0 plus 0 divided by 0. We get LN of 1 divided by 0, which is basically 0 divided by 0. And now this indeterminant form is better for us because we can apply Lapito's rule. Lapita's rule is applicable if we get 0 divided by 0 or infinity divided by infinity. And it tells us that we want to find the derivative of the numerator. And the derivative of the denominator. Let's go ahead and find each derivative. We're going to get a limit as X approaches 0. In the numerator, we have a composite function. So, first of all, we're differentiating LM with respect to the inner function E to the power of 2 X plus X. To get one divided by. E to the power of 2 x plus X, and according to the chain rule we're multiplying by the derivative of the inner function. So e to the power of 2 x plus x derivative. In the denominator we take the derivative of X which is equal to 1. There's no necessity to divide by one, and we can just focus on the numerator. Now, let's get the derivative of E to the power of 2 X plus X, which leaves us with limit as X approaches 0 off. We are going to get a fraction with e to the power of 2 X plus x in the denominator, and in the numerator we're going to take E to the power of 2 x derivative, which is E to the power of 2 X multiplied by 2 x derivative using the chain rule once again. And the derivative of 2X is basically 2, so we can factor out 2. And the derivative of X is 1, so we get 2 e to the power of 2 x plus 1 in the numerator. And now once again, we're attempting the right substitution, which gives us 2 multiplied by E to the power of 0 + 1. Divided by E to the power of 0 + 0. In the denominator, we get 1 + 0, which is 1. So we just want to focus on the numerator. 2 multiplied by 1 + 1 is equal to 3. So now, this is LN of L and L is our limit. So we want to take the anti log, right? L is going to be equal to E, raise to the power of 3, or basically E cubed. Which corresponds to the answer to the, that's our final answer. Thank you for watching.

Indeterminate Powers and ProductsFind the limits in Exercises 53–68.60. lim (x → 0) (e^x + x)^(1/x)

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Indeterminate Powers and Products
Find the limits in Exercises 53–68.
60. lim (x → 0) (e^x + x)^(1/x)