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What (if exists) the $\lim \limits_{n\to \infty}\frac{(n+1)^n-(n-1)^n}{n^n+2}$?

Should I use the binomial theory in the numerator? Please try to keep it as elementary as possible because we are only in the beginning of the course.

Thanks a lot.

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Try dividing both the numerator and denominator by $n^n$. The numerator will now start with the term$(1+\frac{1}{n})^n$. Does this look familiar? – Tom Cooney Apr 13 '12 at 10:39
Thanks, I got: $\frac{(1+\frac{1}{n})^n-(1-\frac{1}{n})^n}{1+\frac{2}{n^n}}$; is that what you meant? if so does the numerator converge to $e-e^{-1}$? and does the denominator converge to infinity? – Anonymous Apr 13 '12 at 10:51
You have the right answer for the numerator, but check your answer for the denominator again. – Tom Cooney Apr 13 '12 at 10:52
Oops, I meant to write that the denominator converges to 1; if I'm correct then the sequence converges to $e-e^{-1}$. – Anonymous Apr 13 '12 at 10:53

For sake of complete answer, this answer will reflect the discussion in the comments of the question.

$$\begin{align*} \lim_{n\rightarrow\infty}\frac{(n+1)^n-(n-1)^n}{n^n+2}&=\lim_{n\rightarrow\infty}\frac{\left(1+\frac{1}{n}\right)^n-\left(1-\frac{1}{n}\right)^n}{1+\frac{2}{n^n}}\quad\text{(Divide by $n^n$)}\\ &=\lim_{n\rightarrow\infty}\frac{e-e^{-1}}{1}\quad\text{(Use $\left(1+\frac{a}{n}\right)^n=e^a$ for numerator)}\\ &=e-e^{-1} \end{align*}$$

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For simplification, does it help you to consider a general $n$ on the (real) number line? Then adjacent integers can be described as $n-1$ and $n+1$ and we have a sequence of three numbers $n-1, n, n+1$. The difference of the two endpoints $n+1 - (n-1)$ is $2$. What then, happens to the differences for powers of these integers?

Does that aid you in any way?

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No it does not. – Did Apr 13 '12 at 10:45
You're not asking the question and I fail to see why you would respond to me. – Autolatry Apr 13 '12 at 11:10
Different approaches guys, don't be sol blind sighted into accepting one answer as the only answer. – Autolatry Apr 13 '12 at 11:46
Fair enough. Care to explain how your post might help the OP? – Did Apr 13 '12 at 11:46

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