# Why $\lim_{n \rightarrow \infty}\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} =5e$?

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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Do you know $\lim_{n\to\infty}(1+(1/n))^n$? It might comes in handy.... – Gerry Myerson Aug 28 '12 at 6:08
The answer is not $5$. – André Nicolas Aug 28 '12 at 6:09
probably 5e ... – Monkey D. Luffy Aug 28 '12 at 6:10
check this out – Monkey D. Luffy Aug 28 '12 at 6:11
Yes, Wolframalpha told me too :) I changed my Edit. – hofmeister Aug 28 '12 at 6:38

$$\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} = \frac{5n^3-3n^2+7}{n(n + 1)^2}\frac{(n+1)^n}{n^n} = \frac{5n^3-3n^2+7}{n(n + 1)^2} \left ( 1 + \frac{1}{n} \right )^n$$
@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 6:57