L’Hôpital’s Rule
Find the limits in Exercises 103–110.

Question

L’Hôpital’s Rule

Find the limits in Exercises 103–110.

108. lim(x→∞)(e^x arctan(e^x))/(e^(2x)+x)

Accepted Answer

Hello. In this video, we are going to be evaluating the given limit by using Laholotal's rule. The limit given to us is the limit as X approaches infinity. Of pi divided by 2 minus tangent inverse of e to the power of 2 x divided by e to the power of -2 x. Now the first thing we want to do is we want to check to see if we have any indeterminate forms. If we were to plug in the limit into the given function, the limit of e to the power of 2 of 2 times infinity will just give us infinity, and the value of tangent inverse at infinity is going to be pi divided by 2. So by plugging in our limit, we get a numerator of pi divided by 2 minus pi divided by 2. And in the denominator e to the power of -2 multiplied by infinity gives us the value of 0. Now further simplifying this will leave us with 0 divided by 0, and this is what we call an indeterminate form. So what we want to do is you want to apply Lahopito's rule in order to simplify the limit. Now the Hopito's rule tells us that we want to take the derivative of the numerator and denominator of a rational function independently. Then once we complete this derivatives, we will go ahead and apply the limit once more and if we get another indeterminate form we will just repeat the process. So, let's go ahead and start by taking the derivative of our numerator. Now again, the numerator is pi divided by 2 minus tangent inverse of e to the power of 2 X. Now the derivative of a constant value pi divided by 2 is just going to be zero. So the derivative of the first term will just be zero. But how do we take the derivative of the second term? While recall that the derivative. Of tangent inverse Of a function of u is defined as u. Divided by 1 plus u squad here u is going to equal to e to the power of 2 x. Now the derivative of E to the power of 2x is going to be 2e the power of 2X. So this tells us that the derivative of the second term is going to be -1 multiplied by 2e to the power of 2 x divided by 1 plus e to the power of 2 x 2, which is e to the power of 4 x. So this is going to be the derivative of our numerator. But what about the denominator? While in the denominator we are just taking the derivative of the exponential value e to the power of -2 x and the derivative of this exponential value. Will be -2 e to the power of 2 x. Now let's go ahead and place these two values back into the original limit. So that will allow us to rewrite the limit as the limit as x approaches infinity of -2 e to the power of 2 x divided by 1 plus e to the power of 4 x, and this will all be divided by -2 e to the power of -2 x. Now we can further simplify this limit by simplifying the complex fraction. And this limit is going to simplify to be the limit as x approaches infinity of e to the power of 4 x divided by 1 plus e to the power of 4 x. Now if we were to apply the limit here, we would end up with infinity divided by infinity. Again, this is an indeterminate form, so we have to apply Lapulpito's law one more time. So again, we will take the derivative of the numerator and denominator independently. The derivative of e to the power of 4 x will be 4 e to the power of 4 x. The derivative of 1 will be 0, and we are taking the derivative of e to the power of 4 x again, which is going to be 4 e to the power of 4 x. So this is going to be our new limit, but notice that the numerator and denominator simplified to be just one, so we are left with the limit as x approaches infinity of just 1. And the limit of any constant value is the constant value itself, so the limit as x approaching infinity of 1 is just going to be 1. This is going to be the value of the given limit by using Lahopito's rule, and with that being said, the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

L’Hôpital’s RuleFind the limits in Exercises 103–110.108. lim(x→∞)(e^x arctan(e^x))/(e^(2x)+x)

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L’Hôpital’s Rule
Find the limits in Exercises 103–110.
108. lim(x→∞)(e^x arctan(e^x))/(e^(2x)+x)