1. ## Limit Question

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!

2. Originally Posted by negligible91
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

3. You also can use the sandwich rule...

4. It is easy to note that $\displaystyle e^2 \leqslant \sqrt[n]{{e^{2n} + e^n }} \leqslant e^2 \sqrt[n]{2}$

5. 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

$\displaystyle \chi$ $\displaystyle \sigma$

6. Thank you very much everyone, I ended up having trouble using L'Hospital's Rule to simplify to e^2 but chisigma's post helped me find another way to solve the problem as well as use L'Hopital's. Thanks!