lim as n approaches infinity of (n/(n+1))^2n
limit equals e?
if so why?
No,Originally Posted by Nichelle14
Consider,
$\displaystyle \lim_{n\to\infty}\left( 1+\frac{1}{n} \right)^n=e $
Since, the limit exists, then you can evaluate, $\displaystyle f(x)=1/x$ because it is countinous for $\displaystyle x>0$.
Thus,
$\displaystyle f\left( \lim_{n\to\infty}\left( 1+\frac{1}{n} \right)^n \right) =f(e)$
Thus, you have,
$\displaystyle \lim_{n\to\infty}\left( \frac{n}{n+1} \right)^n=1/e$
Again, since $\displaystyle g(x)=x^2$ is countinous you can evaluate both sides by it thus,
$\displaystyle g\left(\lim_{n\to\infty}\left( \frac{n}{n+1} \right)^n \right) =g(1/e)$
Thus,
$\displaystyle \lim_{n\to\infty}\left( \frac{n}{n+1} \right)^{2n}=1/e^2$