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Determine whether the sequence converges or diverges. If it converges, find the limit.

$ \{ n^2e^{-n}\} $

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in terms that we're looking at is in squared times e to the minus, and it's another way of writing. This is in squared over each of the end, so we should know already that this is going to go to zero because we have polynomial growth in the numerator and we have exponential growth in the denominator. You'd also prove it using low Patel's rule. All right, this will be an easy thing to work with, with for a little bit tells rule. So we're allowed to use low Patel's room when we have something that would give us infinity over infinity or zero over zero if we were to just plug in the value of in here. So we were just to plug in and equals infinity. We would get infinity over infinity here, right? That's something that we're not actually allowed to do because infinity over infinity is indeterminate form. But if we are in that type of situation, we're allowed to use Low Patel's rule, which means that we take the derivative of the top part and we divide by the derivative of the denominator. So we'd get to in, That's the derivative have been squared divided by either the end. That's the derivative of Indian with respect to end. And then we're in the same set up. So we do low Patel's rule again. We get limit as n goes to infinity of two derivative of two and with respect to end is just too. And then the denominator we still hav e the n So now I would have to over infinity, which is zero. So we do converge We converged to zero right And we should know this immediately from looking at it because again, exponential growth is much faster than polynomial growth. So in the end, this either the end is goingto completely dominate over the n squared and the whole day in term is going to end up going to zero.