\begin{array}{l}
\lim _{n\rightarrow \infty }\left( 1+2^{n} +4^{n}\right)^{\frac{1}{n}}
\end{array}
Here is my thinking:
\begin{array}{l}
\lim _{n\rightarrow \infty }\left( 1+2^{n} +4^{n}\right)^{\frac{1}{n}}\rightarrow \lim _{n\rightarrow \infty } e^{\log\left( 1+2^{n} +4^{n}\right)^{\frac{1}{n}}}\\
\\
\lim _{n\rightarrow \infty }\frac{\log\left( 1+2^{n} +4^{n}\right)}{n}\xrightarrow{I'H}\lim _{n\rightarrow \infty }\frac{2^{n} \log 2\ +\ 4^{n} \log 4}{1+2^{n} +4^{n}} =\lim _{n\rightarrow \infty }\frac{2^{n} \log 2\ +2\cdot \ 4^{n} \log 2}{1+2^{n} +4^{n}}\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\xrightarrow{I'H}\lim _{n\rightarrow \infty }\frac{\log 2\left( 2^{n} \log 2+2\cdot 4^{n} \log 4\right)}{2^{n} \log 2+4^{n} \log 4}\\
=\ \lim _{n\rightarrow \infty }\frac{2^{n} \log 2+2^{2n+2} \log 2}{2^{n} +2^{2n+1}}\\
=\ \lim _{n\rightarrow \infty }\frac{\log 2\left( 1+2^{n+2}\right)}{1+2^{n+1}}\\
=\log 2\\
\\
\therefore \lim _{n\rightarrow \infty }\left( 1+2^{n} +4^{n}\right)^{\frac{1}{n}} =2
\end{array}
And I tried to follow this question: \begin{array}{l}1+2^{n} +4^{n} \sim 4^{n} \ \therefore \lim _{n\rightarrow \infty }\sqrt[n]{4^{n}} =4\end{array}
I'm confused about which is right. Thanks for your help
Best Answer
Your error is in your final step $$ \lim_{n\rightarrow\infty}\frac{\ln(2)(1+2^{n+2})}{1+2^{n+1}} = 2\ln(2)=\ln(4) $$ instead of what you said which was $\ln(2)$. You can get this by multiplying the top and bottom by $\frac{1}{2^{n+1}}$ $$ \lim_{n\rightarrow\infty}\frac{\ln(2)(1+2^{n+2})}{1+2^{n+1}}\frac{\frac{1}{2^{n+1}}}{\frac{1}{2^{n+1}}} = \lim_{n\rightarrow\infty}\frac{\ln(2)(\frac{1}{2^{n+1}}+2)}{\frac{1}{2^{n+1}}+1} = 2\ln(2) $$ since $\frac{1}{2^{n+1}}$ goes to $0$.
$e^{\ln(4)}=4$ which is your desired answer