今回は無理式の極限について見ていきます。
(例題1)次の極限値を求めよ。
(1)\(\displaystyle\lim_{n \to \infty}\displaystyle\frac{3n+1}{\sqrt{n^2+1}+n}\)
(2)\(\displaystyle\lim_{n \to \infty}\displaystyle\frac{1}{\sqrt{n^2+2n}-\sqrt{n^2-2n}}\)
(3)\(\displaystyle\lim_{n \to \infty}(3n-\sqrt{9n^2-2n})\)
(4)\(\displaystyle\lim_{n \to \infty}n\left(\sqrt{4+\displaystyle\frac{1}{n}}-2\right)\)
(解答)
(1)
\(\displaystyle\lim_{n \to \infty}\displaystyle\frac{3n+1}{\sqrt{n^2+1}+n}\)
(\(n\ (=\sqrt{n^2})\) で割ると)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{3+\displaystyle\frac{1}{n}}{\sqrt{1+\displaystyle\frac{1}{n^2}}+1}\)
\(=\displaystyle\frac{3+0}{\sqrt{1+0}+1}\)
\(=\displaystyle\frac{3}{2}\)
(2)
\(\displaystyle\lim_{n \to \infty}\displaystyle\frac{1}{\sqrt{n^2+2n}-\sqrt{n^2-2n}}\)
(いきなり\(\sqrt{n^2}\)で割ってもうまくいかないのでまずは有理化をします)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{1}{\sqrt{n^2+2n}-\sqrt{n^2-2n}}\cdot\displaystyle\frac{\sqrt{n^2+2n}+\sqrt{n^2-2n}}{\sqrt{n^2+2n}+\sqrt{n^2-2n}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{\sqrt{n^2+2n}+\sqrt{n^2-2n}}{(n^2+2n)-(n^2-2n)}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{\sqrt{n^2+2n}+\sqrt{n^2-2n}}{4n}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{\sqrt{1+\displaystyle\frac{2}{n}}+\sqrt{1-\displaystyle\frac{2}{n}}}{4}\)
\(=\displaystyle\frac{\sqrt{1}+\sqrt{1}}{4}\)
\(=\displaystyle\frac{1}{2}\)
(3)
(同様に有理化する)
\(\displaystyle\lim_{n \to \infty}(3n-\sqrt{9n^2-2n})\)
\(=\displaystyle\lim_{n \to \infty}(3n-\sqrt{9n^2-2n})\cdot\displaystyle\frac{3n+\sqrt{9n^2-2n}}{3n+\sqrt{9n^2-2n}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{9n^2-(9n^2-2n)}{3n+\sqrt{9n^2-2n}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{2n}{3n+\sqrt{9n^2-2n}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{2}{3+\sqrt{9-\displaystyle\frac{2}{n}}}\)
\(=\displaystyle\frac{2}{3+\sqrt{9}}\)
\(=\displaystyle\frac{1}{3}\)
(4)
\(\displaystyle\lim_{n \to \infty}n\left(\sqrt{4+\displaystyle\frac{1}{n}}-2\right)\)
\(=\displaystyle\lim_{n \to \infty}n\left(\sqrt{4+\displaystyle\frac{1}{n}}-2\right)\cdot\displaystyle\frac{\sqrt{4+\displaystyle\frac{1}{n}}+2}{\sqrt{4+\displaystyle\frac{1}{n}}+2}\)
\(=\displaystyle\lim_{n \to \infty}n\cdot\displaystyle\frac{(4+\displaystyle\frac{1}{n})-4}{\sqrt{4+\displaystyle\frac{1}{n}}+2}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{1}{\sqrt{4+\displaystyle\frac{1}{n}}+2}\)
\(=\displaystyle\frac{1}{\sqrt{4}+2}\)
\(=\displaystyle\frac{1}{4}\)
(例題2)次の極限値を求めよ。
(1)\(\displaystyle\lim_{n \to \infty}\displaystyle\frac{\sqrt{n+5}-\sqrt{n+3}}{\sqrt{n+1}-\sqrt{n}}\)
(2)\(\displaystyle\lim_{n \to \infty}\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right)\)
(3)\(\displaystyle\lim_{n \to \infty}(\sqrt[3]{n^3-n^2}-n)\)
(3)は3次式の因数分解公式 \(x^3-y^3=(x-y)(x^2+xy+y^2)\) で有理化します。
(解答)
(1)
(分母分子どちらも有理化)
\(\displaystyle\lim_{n \to \infty}\displaystyle\frac{\sqrt{n+5}-\sqrt{n+3}}{\sqrt{n+1}-\sqrt{n}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{(\sqrt{n+5}-\sqrt{n+3})(\sqrt{n+5}+\sqrt{n+3})}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}\cdot\displaystyle\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+5}+\sqrt{n+3}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{(n+5)-(n+3)}{(n+1)-n}\cdot\displaystyle\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+5}+\sqrt{n+3}}\)
\(=\displaystyle\lim_{n \to \infty}2\cdot\displaystyle\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+5}+\sqrt{n+3}}\)
\(=\displaystyle\lim_{n \to \infty}2\cdot\displaystyle\frac{\sqrt{1+\displaystyle\frac{1}{n}}+\sqrt{1}}{\sqrt{1+\displaystyle\frac{5}{n}}+\sqrt{1+\displaystyle\frac{3}{n}}}\)
\(=2\)
(2)
\(\displaystyle\lim_{n \to \infty}\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right)\)
\(=\displaystyle\lim_{n \to \infty}\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right)\cdot\displaystyle\frac{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{(n+\sqrt{n})-(n-\sqrt{n})}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{2\sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}\)
(\(\sqrt{n}\)で割って)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{2}{\sqrt{1+\displaystyle\frac{\sqrt{n}}{n}}+\sqrt{1-\displaystyle\frac{\sqrt{n}}{n}}}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{2}{\sqrt{1+\displaystyle\frac{1}{\sqrt{n}}}+\sqrt{1-\displaystyle\frac{1}{\sqrt{n}}}}\)
\(=1\)
(3)
\(\displaystyle\lim_{n \to \infty}(\sqrt[3]{n^3-n^2}-n)\)
\(=\displaystyle\lim_{n \to \infty}(\sqrt[3]{n^3-n^2}-n)\cdot\displaystyle\frac{\sqrt[3]{(n^3-n^2)^2}+\sqrt[3]{n^3-n^2}\cdot n+n^2}{\sqrt[3]{(n^3-n^2)^2}+\sqrt[3]{n^3-n^2}\cdot n+n^2}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{(n^3-n^2)-n^3}{\sqrt[3]{(n^3-n^2)^2}+\sqrt[3]{n^3-n^2}\cdot n+n^2}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{-n^2}{\sqrt[3]{(n^3-n^2)^2}+\sqrt[3]{n^3-n^2}\cdot n+n^2}\)
極限は次数が一番大きいものでほとんど決まるので、最大の次数の項だけを考えると分母の3乗根は
\(\sqrt[3]{(n^3-n^2)^2}≒\sqrt[3]{(n^3)^2}=n^2\)
\(\sqrt[3]{n^3-n^2}\cdot n≒\sqrt[3]{n^3}\cdot n=n^2\)
よって\(n^2\)で割るとうまくいくわけです。(このように累乗根の次数を先に大まかに決定すると見通しがたつ)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{-1}{\sqrt[3]{\displaystyle\frac{1}{n^6}(n^3-n^2)^2}+\sqrt[3]{\displaystyle\frac{1}{n^3}(n^3-n^2)}\cdot 1+1}\)
(\(n^6=(n^3)^2\) より2乗の括弧内に入れます)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{-1}{\sqrt[3]{\left(\displaystyle\frac{n^3-n^2}{n^3}\right)^2}+\sqrt[3]{1-\displaystyle\frac{1}{n}}+1}\)
\(=\displaystyle\lim_{n \to \infty}\displaystyle\frac{-1}{\sqrt[3]{\left(1-\displaystyle\frac{1}{n}\right)^2}+\sqrt[3]{1-\displaystyle\frac{1}{n}}+1}\)
\(=-\displaystyle\frac{1}{3}\)
以上になります。お疲れさまでした。
ここまで見ていただきありがとうございました。
next→和や積の形の極限 back→数列の極限の基礎例題①(分数式など)