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

  • 1
    Do you know $\lim_{n\to\infty}(1+(1/n))^n$? It might comes in handy....2012-08-28
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    The answer is not $5$.2012-08-28
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    probably 5e ...2012-08-28
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    [check this out](http://www.wolframalpha.com/input/?i=\lim_{n+\rightarrow+\infty}\frac{%285n^3-3n^2%2B7%29%28n%2B1%29^n}{n^{n%2B1}%28n%2B1%29^2})2012-08-28
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    Yes, Wolframalpha told me too :) I changed my **Edit**.2012-08-28
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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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    Ok. Thanks. Pleas see my **Edit**. Does I understand it correctly? Greetz.2012-08-28
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    @hofmeister Yes, your edit seems fine. Just be careful that you should add limits in front of your expressions. And check the last limit you wrote, the answer should be $5$, but you wrote $5e$ instead, although I'm sure that was just a typo.2012-08-28
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    Thanks very much. I will keep in my mind. Wish a nice day.2012-08-28