## 問題の内容解析学極限有理化三角関数マクローリン展開2025/7/9## 問題の内容次の5つの極限を計算します。(1) limx→12x+3−4x+1x−1\lim_{x \to 1} \frac{\sqrt{2x+3} - \sqrt{4x+1}}{x-1}limx→1x−12x+3−4x+1(2) limx→∞(4x2+x−2x)\lim_{x \to \infty} (\sqrt{4x^2 + x} - 2x)limx→∞(4x2+x−2x)(3) limx→0sin3xsin5x\lim_{x \to 0} \frac{\sin 3x}{\sin 5x}limx→0sin5xsin3x(4) limx→0(1+3x)2x\lim_{x \to 0} (1+3x)^{\frac{2}{x}}limx→0(1+3x)x2(5) limx→0tan−1x−xx3\lim_{x \to 0} \frac{\tan^{-1}x - x}{x^3}limx→0x3tan−1x−x## 解き方の手順(1) limx→12x+3−4x+1x−1\lim_{x \to 1} \frac{\sqrt{2x+3} - \sqrt{4x+1}}{x-1}limx→1x−12x+3−4x+1分母と分子に 2x+3+4x+1\sqrt{2x+3} + \sqrt{4x+1}2x+3+4x+1 を掛けて、有理化します。limx→1(2x+3−4x+1)(2x+3+4x+1)(x−1)(2x+3+4x+1)=limx→1(2x+3)−(4x+1)(x−1)(2x+3+4x+1)=limx→1−2x+2(x−1)(2x+3+4x+1)\lim_{x \to 1} \frac{(\sqrt{2x+3} - \sqrt{4x+1})(\sqrt{2x+3} + \sqrt{4x+1})}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})} = \lim_{x \to 1} \frac{(2x+3) - (4x+1)}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})} = \lim_{x \to 1} \frac{-2x+2}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})}limx→1(x−1)(2x+3+4x+1)(2x+3−4x+1)(2x+3+4x+1)=limx→1(x−1)(2x+3+4x+1)(2x+3)−(4x+1)=limx→1(x−1)(2x+3+4x+1)−2x+2=limx→1−2(x−1)(x−1)(2x+3+4x+1)=limx→1−22x+3+4x+1= \lim_{x \to 1} \frac{-2(x-1)}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})} = \lim_{x \to 1} \frac{-2}{\sqrt{2x+3} + \sqrt{4x+1}}=limx→1(x−1)(2x+3+4x+1)−2(x−1)=limx→12x+3+4x+1−2x=1x = 1x=1 を代入すると、−25+5=−225=−15\frac{-2}{\sqrt{5} + \sqrt{5}} = \frac{-2}{2\sqrt{5}} = -\frac{1}{\sqrt{5}}5+5−2=25−2=−51(2) limx→∞(4x2+x−2x)\lim_{x \to \infty} (\sqrt{4x^2 + x} - 2x)limx→∞(4x2+x−2x)分母と分子に 4x2+x+2x\sqrt{4x^2 + x} + 2x4x2+x+2x を掛けて、有理化します。limx→∞(4x2+x−2x)(4x2+x+2x)4x2+x+2x=limx→∞(4x2+x)−4x24x2+x+2x=limx→∞x4x2+x+2x\lim_{x \to \infty} \frac{(\sqrt{4x^2 + x} - 2x)(\sqrt{4x^2 + x} + 2x)}{\sqrt{4x^2 + x} + 2x} = \lim_{x \to \infty} \frac{(4x^2 + x) - 4x^2}{\sqrt{4x^2 + x} + 2x} = \lim_{x \to \infty} \frac{x}{\sqrt{4x^2 + x} + 2x}limx→∞4x2+x+2x(4x2+x−2x)(4x2+x+2x)=limx→∞4x2+x+2x(4x2+x)−4x2=limx→∞4x2+x+2xx分子と分母を xxx で割ります。limx→∞14+1x+2=14+0+2=12+2=14\lim_{x \to \infty} \frac{1}{\sqrt{4 + \frac{1}{x}} + 2} = \frac{1}{\sqrt{4+0} + 2} = \frac{1}{2+2} = \frac{1}{4}limx→∞4+x1+21=4+0+21=2+21=41(3) limx→0sin3xsin5x\lim_{x \to 0} \frac{\sin 3x}{\sin 5x}limx→0sin5xsin3xlimx→0sin3xsin5x=limx→0sin3x3x⋅5xsin5x⋅3x5x=limx→0sin3x3x⋅limx→05xsin5x⋅limx→03x5x\lim_{x \to 0} \frac{\sin 3x}{\sin 5x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{5x}{\sin 5x} \cdot \frac{3x}{5x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \lim_{x \to 0} \frac{5x}{\sin 5x} \cdot \lim_{x \to 0} \frac{3x}{5x}limx→0sin5xsin3x=limx→03xsin3x⋅sin5x5x⋅5x3x=limx→03xsin3x⋅limx→0sin5x5x⋅limx→05x3xlimx→0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1limx→0xsinx=1 であるから、=1⋅1⋅35=35= 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}=1⋅1⋅53=53(4) limx→0(1+3x)2x\lim_{x \to 0} (1+3x)^{\frac{2}{x}}limx→0(1+3x)x2y=(1+3x)2xy = (1+3x)^{\frac{2}{x}}y=(1+3x)x2 とおくと、 lny=2xln(1+3x)\ln y = \frac{2}{x} \ln(1+3x)lny=x2ln(1+3x)limx→0lny=limx→02ln(1+3x)x=limx→02ln(1+3x)3x⋅3=2⋅1⋅3=6\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{2 \ln(1+3x)}{x} = \lim_{x \to 0} \frac{2 \ln(1+3x)}{3x} \cdot 3 = 2 \cdot 1 \cdot 3 = 6limx→0lny=limx→0x2ln(1+3x)=limx→03x2ln(1+3x)⋅3=2⋅1⋅3=6limx→0ln(1+x)x=1\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1limx→0xln(1+x)=1 であるから、limx→0lny=6\lim_{x \to 0} \ln y = 6limx→0lny=6よって、 limx→0y=e6\lim_{x \to 0} y = e^6limx→0y=e6(5) limx→0tan−1x−xx3\lim_{x \to 0} \frac{\tan^{-1}x - x}{x^3}limx→0x3tan−1x−xtan−1x\tan^{-1}xtan−1x のマクローリン展開を利用します。tan−1x=x−x33+x55−x77+⋯\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdotstan−1x=x−3x3+5x5−7x7+⋯limx→0tan−1x−xx3=limx→0(x−x33+x55−⋯ )−xx3=limx→0−x33+x55−⋯x3=limx→0(−13+x25−⋯ )\lim_{x \to 0} \frac{\tan^{-1}x - x}{x^3} = \lim_{x \to 0} \frac{(x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots) - x}{x^3} = \lim_{x \to 0} \frac{-\frac{x^3}{3} + \frac{x^5}{5} - \cdots}{x^3} = \lim_{x \to 0} (-\frac{1}{3} + \frac{x^2}{5} - \cdots)limx→0x3tan−1x−x=limx→0x3(x−3x3+5x5−⋯)−x=limx→0x3−3x3+5x5−⋯=limx→0(−31+5x2−⋯)=−13= -\frac{1}{3}=−31## 最終的な答え(1) −15-\frac{1}{\sqrt{5}}−51(2) 14\frac{1}{4}41(3) 35\frac{3}{5}53(4) e6e^6e6(5) −13-\frac{1}{3}−31