자연계열 2-1번

$\displaystyle a=1$일 때, $\displaystyle \lim_{n\to\infty}\frac{1}{n}\log(a^{2n}+a^{5n})=\lim_{n\to\infty}\frac{1}{n}\log2=0$이다.

$\displaystyle 0<a<1$일 때, $\displaystyle \begin{aligned}\lim_{n\to\infty}\frac{1}{n}\log(a^{2n}+a^{5n})&=\lim_{n\to\infty}\frac{1}{n}\log a^{2n}(1+a^{3n})=\lim_{n\to\infty}\frac{1}{n}[\log a^{2n}+\log(1+a^{3n})]\\&=\lim_{n\to\infty}\left[2\log a+\frac{\log(1+a^{3n})}{n}\right]=2\log a\end{aligned}$

$\displaystyle 1<a<\infty$일 때, $\displaystyle \begin{aligned}\lim_{n\to\infty}\frac{1}{n}\log(a^{2n}+a^{5n})&=\lim_{n\to\infty}\frac{1}{n}\log a^{5n}(a^{-3n}+1)=\lim_{n\to\infty}\frac{1}{n}[\log a^{5n}+\log(a^{-3n}+1)]\\&=\lim_{n\to\infty}\left[5\log a+\frac{\log(a^{-3n}+1)}{n}\right]=5\log a\end{aligned}$