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# Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\displaystyle \lim_{x\to \infty} x^{e^{-x}}$

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the first problem, we're going to be taking the limit as X goes to infinity of the function. Um So in order to do this one, we're gonna want to take the natural log of both sides. And when we do that we can simplify it to be the limit as X approaches infinity of the natural log effect over E to the X. The reason why we do this is because now it makes it much easier to evaluate the limit. When we plug in X going to infinity, we get infinity over infinity. That's an indeterminate form though. So we're going to need to use local tells rule the derivative of natural log of X is one over X. And the derivative of E to the X is just to the X. Now, when we plug in infinity we will get here. One over infinity is zero. So it's got zero and then each of the zero is one, Even the Zeros one. So we get one right here. But zero is not our answer. In fact, the natural log of Y equals zero is our answer, because we took the natural log of both sides. Now, when we solve for why we see that Y equals E. To the zero, which equals one, so one is going to be our final answer.

California Baptist University

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