Question

[(n-3)/(n-1)]^n+1 ?! -> lim=(1+1/n)^n

Answer 1

the second limit is e if you are taking the limit as n goes to infinity

Answer 2

yep it's e but solve the first one to this limit e..?!

Answer 3

no i think it is
\[e^{-2}\]

Answer 4

teh result is \[e^{4}\]

Answer 5

let L = the limit...
\[L = \lim_{n \rightarrow \infty} \left(\frac{n-3}{n-1}\right)^{n+1} \]
then
\[ \ln(L) =\lim_{n \rightarrow \infty} (n+1)\ln(\frac{n-3}{n-1}) \]
which can be written
\[ \lim_{n\rightarrow \infty} \frac{\ln(\frac{n-3}{n-1})}{\frac{1}{n+1}} \]
We can now apply l'Hospital's rule...
\[ \ln(L) = \lim_{n\rightarrow \infty} \frac{ \frac{n-1}{n-3} \cdot \frac{2}{(n-1)^2}}{\frac{-1}{(n+1)^2}} = \lim_{n\rightarrow \infty}-2 \left(\frac{n+1}{n-1}\right)^2 \left(\frac{n-1}{n-3}\right) = -2\]
so from there it follows that
\[ L = e^{-2} \]

Answer 6

could someone to solve with the result \[e^{4}\] ?

Answer 7

No, because that's not the answer...

Answer 8

You're right, thanks.