1. ## limit

I cant calculate this limit correctly...

thank you very much for your help

2. $\lim_{n\to\infty}(1-(\frac{2}{n+2})^{\frac{n+2}{n}}) = \lim_{n\to\infty}(1-(0)^1) = 1$

you answered for the limit of one multiplier...but you got n multipliers...so its not that simple...but thanks anyway

4. nevemind I solved it...

the limit is $e^-2$ by the way

5. I realized that but upon inspection, every exponent should go to 0 as n approaches infinity, regardless of the number inside the brackets every multiplier will equal one. Try graphing a few terms, the limit is one. e^-2 is approximately 0.135 it's much too low.

6. Originally Posted by UriK
nevemind I solved it...

the limit is $e^{-2}$ by the way
I'm inclined to agree. We have here

$\lim_{n\to\infty}\prod_{k=1}^n\left(1-\frac{2}{k+2}\right)^{\frac{k+2}{n}}=\lim_{n\to\in fty}\prod_{k=1}^n\left[\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}$ $=\lim_{n\to\infty}\left[\prod_{k=1}^n\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}$

Fact: $\left(1-\frac{1}{\alpha}\right)^{\alpha}$ tends to $e^{-1}$ as $\alpha$ gets large. (In this case $\alpha=\frac{k+2}{2}$.) Now the argument gets a bit informal (but hopefully stays correct), but as $n$ gets very large, the product becomes approximately $e^{-1}\cdot e^{-1}\cdot...e^{-1}=e^{-n}$. So we have:

$\lim_{n\to\infty}\left[\prod_{k=1}^n\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}=\lim_{n\to\infty}(e^{-n})^{2/n}=\boxed{e^{-2}}$