$\lim_{x\to 0} \frac{\sqrt{3x+1}-1}{\sqrt{6x+1}-1}$ を計算します。解析学極限関数の極限有理化2025/5/121. 問題の内容limx→03x+1−16x+1−1\lim_{x\to 0} \frac{\sqrt{3x+1}-1}{\sqrt{6x+1}-1}limx→06x+1−13x+1−1 を計算します。2. 解き方の手順この極限を計算するために、分子と分母をそれぞれ有利化します。まず、分子を有利化します。分子と分母に 3x+1+1\sqrt{3x+1}+13x+1+1 をかけます。3x+1−16x+1−1=(3x+1−1)(3x+1+1)(6x+1−1)(3x+1+1)=(3x+1)−1(6x+1−1)(3x+1+1)=3x(6x+1−1)(3x+1+1)\frac{\sqrt{3x+1}-1}{\sqrt{6x+1}-1} = \frac{(\sqrt{3x+1}-1)(\sqrt{3x+1}+1)}{(\sqrt{6x+1}-1)(\sqrt{3x+1}+1)} = \frac{(3x+1)-1}{(\sqrt{6x+1}-1)(\sqrt{3x+1}+1)} = \frac{3x}{(\sqrt{6x+1}-1)(\sqrt{3x+1}+1)}6x+1−13x+1−1=(6x+1−1)(3x+1+1)(3x+1−1)(3x+1+1)=(6x+1−1)(3x+1+1)(3x+1)−1=(6x+1−1)(3x+1+1)3x次に、分母を有利化します。6x+1+1\sqrt{6x+1}+16x+1+1 を分子と分母にかけます。3x(6x+1−1)(3x+1+1)=3x(6x+1+1)(6x+1−1)(6x+1+1)(3x+1+1)=3x(6x+1+1)((6x+1)−1)(3x+1+1)=3x(6x+1+1)6x(3x+1+1)\frac{3x}{(\sqrt{6x+1}-1)(\sqrt{3x+1}+1)} = \frac{3x(\sqrt{6x+1}+1)}{(\sqrt{6x+1}-1)(\sqrt{6x+1}+1)(\sqrt{3x+1}+1)} = \frac{3x(\sqrt{6x+1}+1)}{((6x+1)-1)(\sqrt{3x+1}+1)} = \frac{3x(\sqrt{6x+1}+1)}{6x(\sqrt{3x+1}+1)}(6x+1−1)(3x+1+1)3x=(6x+1−1)(6x+1+1)(3x+1+1)3x(6x+1+1)=((6x+1)−1)(3x+1+1)3x(6x+1+1)=6x(3x+1+1)3x(6x+1+1)x≠0x \neq 0x=0 のとき、3x(6x+1+1)6x(3x+1+1)=3(6x+1+1)6(3x+1+1)=6x+1+12(3x+1+1)\frac{3x(\sqrt{6x+1}+1)}{6x(\sqrt{3x+1}+1)} = \frac{3(\sqrt{6x+1}+1)}{6(\sqrt{3x+1}+1)} = \frac{\sqrt{6x+1}+1}{2(\sqrt{3x+1}+1)}6x(3x+1+1)3x(6x+1+1)=6(3x+1+1)3(6x+1+1)=2(3x+1+1)6x+1+1したがって、limx→03x+1−16x+1−1=limx→06x+1+12(3x+1+1)=6(0)+1+12(3(0)+1+1)=1+12(1+1)=1+12(1+1)=22(2)=24=12\lim_{x\to 0} \frac{\sqrt{3x+1}-1}{\sqrt{6x+1}-1} = \lim_{x\to 0} \frac{\sqrt{6x+1}+1}{2(\sqrt{3x+1}+1)} = \frac{\sqrt{6(0)+1}+1}{2(\sqrt{3(0)+1}+1)} = \frac{\sqrt{1}+1}{2(\sqrt{1}+1)} = \frac{1+1}{2(1+1)} = \frac{2}{2(2)} = \frac{2}{4} = \frac{1}{2}limx→06x+1−13x+1−1=limx→02(3x+1+1)6x+1+1=2(3(0)+1+1)6(0)+1+1=2(1+1)1+1=2(1+1)1+1=2(2)2=42=213. 最終的な答え12\frac{1}{2}21