1.次の数列の極限を求めなさい。
(1)\(\displaystyle \lim_{n\to\infty}(2n^3-7n^2-9n-4)\)
\(\displaystyle =\lim_{n\to\infty}n^3\left(2-\frac{7}{n}-\frac{9}{n^2}-\frac{4}{n^3}\right)\)
\(\displaystyle =\infty\)
(2)\(\displaystyle \lim_{n\to\infty}\frac{3n+2}{2n+1}\)
\(\displaystyle =\lim_{n\to\infty}\frac{3+\frac{2}{n}}{2+\frac{1}{n}}\)
\(\displaystyle =\frac{3}{2}\)
(3)\(\displaystyle \lim_{n\to\infty}\frac{\sqrt{2n^2+3n+5}}{n+1}\)
\(\displaystyle =\lim_{n\to\infty}\frac{\sqrt{2+\frac{3}{n}+\frac{5}{n^2}}}{1+\frac{1}{n}}\)
\(\displaystyle =\sqrt{2}\)
(4)\(\displaystyle \lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n})\)
\(\displaystyle =\lim_{n\to\infty}\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
\(\displaystyle =0\)
(5)\(\displaystyle \lim_{n\to\infty}\sqrt{n}(\sqrt{n+1}-\sqrt{n})\)
\(\displaystyle =\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}\)
\(\displaystyle =\frac{1}{2}\)
2.数列\(\{a_n\}^\infty_{n=1}\)が\(\displaystyle \lim_{n\to\infty}(2n+3)a_n=5\)をみたすとき、次の数列の極限を求めなさい。
(1)\(\displaystyle \lim_{n\to\infty}a_n\)
\(\displaystyle b_n=(2n+3)a_n\)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、
\(\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{b_n}{2n+3}=\frac{5}{\infty}=0\)
(2)\(\displaystyle \lim_{n\to\infty}na_n\)
\(\displaystyle b_n=(2n+3)a_n\)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、
\(\displaystyle \lim_{n\to\infty}na_n=\lim_{n\to\infty}\frac{nb_n}{2n+3}=\lim_{n\to\infty}\frac{b_n}{2+\frac{3}{n}}=\frac{5}{2}\)