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I have a small exercise and I don’t know who to get the result.

The exercise is: $ \lim_{n \rightarrow \infty}\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} $

I did following transformations: $ \frac{(5n^3-3n^2+7)(n+1)^{n-2}}{n^{n+1}} \\ (5n^{2-n}-3n^{1-n}+7^{-1-n})(n+1)^{n-2} \\ (\frac{5}{n^{-2+n}} - \frac{3}{n^{-1+n}} + \frac{7}{n^{1+n}})(n+1)^{n-2} $

But none of them helped me to see the result. It would be great if someone could explain it to me.

Edit

@adrian-barquero Ok. Fist you factories $^n$ and get $ \frac{(n+1)^n}{n^n} = (1+\frac{1}{n})^n = e \\ $

In the other fraction I could extend with $n^3$ $ \frac{5n^3-3n^2+7}{n(n + 1)^2} = \frac{n^3(5 - \frac{3n^2}{n^3} + \frac{7}{n^3})}{n^3(1 + \frac{2n^2}{n^3} + \frac{n}{n^3})} = 5 $

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    What does "Fist you factories ${}^n$" mean?2012-08-28

1 Answers 1

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I would recommend rearranging as

$ \frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} = \frac{5n^3-3n^2+7}{n(n + 1)^2}\frac{(n+1)^n}{n^n} = \frac{5n^3-3n^2+7}{n(n + 1)^2} \left ( 1 + \frac{1}{n} \right )^n $

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    Thanks very much. I will keep in my mind. Wish a nice day.2012-08-28