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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 Answers1

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I would recommend rearranging as

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

Adrián Barquero
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  • Ok. Thanks. Pleas see my Edit. Does I understand it correctly? Greetz. –  Aug 28 '12 at 06:31
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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. – Adrián Barquero Aug 28 '12 at 06:57
  • Thanks very much. I will keep in my mind. Wish a nice day. –  Aug 28 '12 at 06:58