limit as n-> infinity
of
(e^(2n)+e^(n))^(1/n)
Apparently the answer is e^2 but I'm completely clueless how to solve other than probably using ln. Any help would be nice, thank you!
$\displaystyle \displaystyle{\left(e^{2n}+e^n\right)^{1/n}=e^{\frac{1}{n}\ln(e^{2n}+e^n)}}$ , and now:
$\displaystyle \lim\limits_{n\rightarrow\infty}\frac{\ln(e^{2n}+e ^n)}{n}=2$ , as you can easily check by applying L'Hospital to $\displaystyle \lim\limits_{x\rightarrow\infty}\frac{\ln(e^{2x}+e ^x)}{x}$
Tonio
Is...
$\displaystyle \displaystyle \ln \{(e^{2n} + e^{n})^{\frac{1}{n}}\}= \frac{1}{n}\ \ln \{e^{2n}\ (1+e^{-n})\} = 2 + \frac{1}{n}\ \ln (1+e^{-n}) $ (1)
Now You have to compute the limit of (1) if n tends to infinity...
Merry Christmas from Italy
$\displaystyle \chi$ $\displaystyle \sigma$