与えられた8つの極限値を求めます。 (1) $\lim_{n \to \infty} \frac{\sqrt{2n+1}}{\sqrt{n}}$ (2) $\lim_{n \to \infty} \frac{4n}{\sqrt{n^2+2n}+n}$ (3) $\lim_{n \to \infty} (\sqrt{n+3}-\sqrt{n})$ (4) $\lim_{n \to \infty} (\sqrt{n^2+1}-n)$ (5) $\lim_{n \to \infty} \frac{1}{\sqrt{n+2}-\sqrt{n}}$ (6) $\lim_{n \to \infty} \frac{3}{\sqrt{n^2+3n}-n}$ (7) $\lim_{n \to \infty} (\sqrt{n^2+4n}-n)$ (8) $\lim_{n \to \infty} \sqrt{n+1}(\sqrt{n+2}-\sqrt{n-1})$

## 問題の解答

1. 問題の内容

与えられた8つの極限値を求めます。

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

(2) $\lim_{n \to \infty} \frac{4n}{\sqrt{n^2+2n}+n}$

(3) $\lim_{n \to \infty} (\sqrt{n+3}-\sqrt{n})$

(4) $\lim_{n \to \infty} (\sqrt{n^2+1}-n)$

(5) $\lim_{n \to \infty} \frac{1}{\sqrt{n+2}-\sqrt{n}}$

(6) $\lim_{n \to \infty} \frac{3}{\sqrt{n^2+3n}-n}$

(7) $\lim_{n \to \infty} (\sqrt{n^2+4n}-n)$

(8) $\lim_{n \to \infty} \sqrt{n+1}(\sqrt{n+2}-\sqrt{n-1})$

2. 解き方の手順

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

(2) $\lim_{n \to \infty} \frac{4n}{\sqrt{n^2+2n}+n} = \lim_{n \to \infty} \frac{4n}{n\sqrt{1+\frac{2}{n}}+n} = \lim_{n \to \infty} \frac{4}{\sqrt{1+\frac{2}{n}}+1} = \frac{4}{\sqrt{1+0}+1} = \frac{4}{2} = 2$

(3) $\lim_{n \to \infty} (\sqrt{n+3}-\sqrt{n}) = \lim_{n \to \infty} \frac{(\sqrt{n+3}-\sqrt{n})(\sqrt{n+3}+\sqrt{n})}{\sqrt{n+3}+\sqrt{n}} = \lim_{n \to \infty} \frac{(n+3)-n}{\sqrt{n+3}+\sqrt{n}} = \lim_{n \to \infty} \frac{3}{\sqrt{n+3}+\sqrt{n}} = 0$

(4) $\lim_{n \to \infty} (\sqrt{n^2+1}-n) = \lim_{n \to \infty} \frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{\sqrt{n^2+1}+n} = \lim_{n \to \infty} \frac{(n^2+1)-n^2}{\sqrt{n^2+1}+n} = \lim_{n \to \infty} \frac{1}{\sqrt{n^2+1}+n} = 0$

(5) $\lim_{n \to \infty} \frac{1}{\sqrt{n+2}-\sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n+2}+\sqrt{n}}{(\sqrt{n+2}-\sqrt{n})(\sqrt{n+2}+\sqrt{n})} = \lim_{n \to \infty} \frac{\sqrt{n+2}+\sqrt{n}}{(n+2)-n} = \lim_{n \to \infty} \frac{\sqrt{n+2}+\sqrt{n}}{2} = \infty$

(6) $\lim_{n \to \infty} \frac{3}{\sqrt{n^2+3n}-n} = \lim_{n \to \infty} \frac{3(\sqrt{n^2+3n}+n)}{(\sqrt{n^2+3n}-n)(\sqrt{n^2+3n}+n)} = \lim_{n \to \infty} \frac{3(\sqrt{n^2+3n}+n)}{(n^2+3n)-n^2} = \lim_{n \to \infty} \frac{3(\sqrt{n^2+3n}+n)}{3n} = \lim_{n \to \infty} \frac{\sqrt{n^2+3n}+n}{n} = \lim_{n \to \infty} \frac{n\sqrt{1+\frac{3}{n}}+n}{n} = \lim_{n \to \infty} (\sqrt{1+\frac{3}{n}}+1) = \sqrt{1+0}+1 = 2$

(7) $\lim_{n \to \infty} (\sqrt{n^2+4n}-n) = \lim_{n \to \infty} \frac{(\sqrt{n^2+4n}-n)(\sqrt{n^2+4n}+n)}{\sqrt{n^2+4n}+n} = \lim_{n \to \infty} \frac{(n^2+4n)-n^2}{\sqrt{n^2+4n}+n} = \lim_{n \to \infty} \frac{4n}{\sqrt{n^2+4n}+n} = \lim_{n \to \infty} \frac{4n}{n\sqrt{1+\frac{4}{n}}+n} = \lim_{n \to \infty} \frac{4}{\sqrt{1+\frac{4}{n}}+1} = \frac{4}{\sqrt{1+0}+1} = \frac{4}{2} = 2$

(8) $\lim_{n \to \infty} \sqrt{n+1}(\sqrt{n+2}-\sqrt{n-1}) = \lim_{n \to \infty} \sqrt{n+1} \frac{(\sqrt{n+2}-\sqrt{n-1})(\sqrt{n+2}+\sqrt{n-1})}{\sqrt{n+2}+\sqrt{n-1}} = \lim_{n \to \infty} \sqrt{n+1} \frac{(n+2)-(n-1)}{\sqrt{n+2}+\sqrt{n-1}} = \lim_{n \to \infty} \sqrt{n+1} \frac{3}{\sqrt{n+2}+\sqrt{n-1}} = \lim_{n \to \infty} \frac{3\sqrt{n+1}}{\sqrt{n+2}+\sqrt{n-1}} = \lim_{n \to \infty} \frac{3\sqrt{n(1+\frac{1}{n})}}{\sqrt{n(1+\frac{2}{n})}+\sqrt{n(1-\frac{1}{n})}} = \lim_{n \to \infty} \frac{3\sqrt{n}\sqrt{1+\frac{1}{n}}}{\sqrt{n}\sqrt{1+\frac{2}{n}}+\sqrt{n}\sqrt{1-\frac{1}{n}}} = \lim_{n \to \infty} \frac{3\sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{2}{n}}+\sqrt{1-\frac{1}{n}}} = \frac{3\sqrt{1+0}}{\sqrt{1+0}+\sqrt{1-0}} = \frac{3}{1+1} = \frac{3}{2}$

3. 最終的な答え

(1) $\sqrt{2}$

(2) 2

(3) 0

(4) 0

(5) $\infty$

(6) 2

(7) 2

(8) $\frac{3}{2}$