86. This exercise explores the difference between
lim(x→...

Question

86. This exercise explores the difference between

lim(x→∞)(1 + 1/x²)^x

and

lim(x→∞)(1 + 1/x)^x = e

c. Confirm your estimate of lim(x→∞)f(x) by calculating it with l’Hôpital’s Rule.

Accepted Answer

Welcome back everyone. Find the limit as x approaches infinity of 1 + 2 divided by x raised to the power of x. A says 1, B E C E 2, and d e to the power of 1/22. So for this problem, let's begin by rewriting the limit. Let's suppose that our limit is equal tol and L is equal to the limit as x approaches infinity of 1 + 2 divided by x raised to the power of x. Let's understand the behavior of the functions x approaches infinity. That's 1 + 2 divided by infinity. Raise to the power of infinity, right? So now 2 divided by infinity approaches. 0 and we simply get 1 raise the power of infinity, which is an indeterminate form. For this indeterminate form, typically we once take ln of l. So let's go ahead and do that. Ln of l is equal to limit as x approaches infinity of ln of 1 + 2 divided by x raised the power of x. And now what we can do is simply use the properties of logarithms. So that would be a limit as x approaches infinity of x multiplied by ln of 1 + 2 divided by x, and here we are simply applying the power rule of logarithms. Now let's go ahead and perform substitution again. This is going to be infinity multiplied by ln of 1 + 2 divided by infinity, or basically ln of 1, which is 0, right? So now we get infinity multiplied by 0. That's again an indeterminate form. What we can do is simply rearrange this expression so that we can apply Lapel's rule, and one of the ways is to go ahead and rewrite multiplication in terms of division. So we're going to leave ln of 1 + 2 divided by x in the numerator, and instead of multiplying by x, we're going to divide by the reciprocal of x, which is 1 divided by x. And that's the same thing, right? Essentially dividing by 1. Divided by x is multiplying by x. But what we can notice is that now we get 0 divided by 0, which is an indeterminate form that is suitable for Lapito's rule. So let's go ahead and apply Lapito's rule. We're going to get the limit as x approaches infinity. What we want to do is simply take the derivative of the numerator and the denominator. So starting with the numerator, we're going to notice that this is a composite function. We're differentiating the natural log with respect to 1 + 2 divided by x, which is 1. Divided by 1 + 2 divided by x and we want to multiply by the derivative of the inner function which is 1 + 2 divided by x, right. And we want to differentiate the denominator, so we want to find the derivative of 1 divided by x. Now, let's go ahead and simplify. This is going to be a limit as x approaches infinity. Of 1 divided by 1 + 2 divided by x multiplied by the derivative of 1 is 0, and what we can do is simply rewrite the derivative of 2 divided by x as 2 multiplied by 1 divided by x derivative, we can see that, right, because according to the properties of derivatives, we can always factor out the constant. And we are dividing by the derivative of 1 divided by x. So notice now that these derivatives can be canceled out. And from here we can compute the limit. So that'd be 1 divided by 1 plus 2 divided by infinity. The final term approach is 0, so we basically get 1. Multiplied by 2. And 1 multiplied by 2 is simply 2. Well done. So we know that ln of the limit is equal to 2. So the limit l itself is going to be the anti log that's e raised to the power of 2 or basically e squared, which corresponds to the answer of choice c. Thank you for watching.

86. This exercise explores the difference betweenlim(x→∞)(1 + 1/x²)^xandlim(x→∞)(1 + 1/x)^x = ec. Confirm your estimate of lim(x→∞)f(x) by calculating it with l’Hôpital’s Rule.

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86. This exercise explores the difference between
lim(x→∞)(1 + 1/x²)^x
and
lim(x→∞)(1 + 1/x)^x = e
c. Confirm your estimate of lim(x→∞)f(x) by calculating it with l’Hôpital’s Rule.