【微分積分】1-1-2 数列の極限の性質｜問題集

\(\displaystyle =\lim_{n\to\infty}n^3\left(2-\frac{7}{n}-\frac{9}{n^2}-\frac{4}{n^3}\right)\)
\(=\infty\)

\(\displaystyle =\lim_{n\to\infty}\frac{3+\frac{2}{n}}{2+\frac{1}{n}}\)
\(\displaystyle =\frac{3}{2}\)

\(\displaystyle =\lim_{n\to\infty}\frac{\sqrt{2+\frac{3}{n}+\frac{5}{n^2}}}{1+\frac{1}{n}}\)
\(=\sqrt{2}\)

\(\displaystyle =\lim_{n\to\infty}\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
\(=0\)

\(\displaystyle =\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}\)
\(\displaystyle =\frac{1}{2}\)

\(\displaystyle b_n=(2n+3)a_n\)とおくと、
\(\displaystyle \lim_{n\to\infty}b_n=5\)
また、
\(\displaystyle a_n=\frac{b_n}{2n+3}\)
よって、
\(\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{b_n}{2n+3}=\frac{5}{\infty}=0\)

\(\displaystyle b_n=(2n+3)a_n\)とおくと、
\(\displaystyle \lim_{n\to\infty}b_n=5\)
また、
\(\displaystyle a_n=\frac{b_n}{2n+3}\)
よって、
\(\displaystyle \lim_{n\to\infty}na_n=\lim_{n\to\infty}\frac{nb_n}{2n+3}=\lim_{n\to\infty}\frac{b_n}{2+\frac{3}{n}}=\frac{5}{2}\)