与えられた極限を計算します。 $$\lim_{x \to 1} \left( \frac{x}{1-x} - \frac{1}{\log x} \right) (x-1)$$解析学極限テイラー展開ロピタルの定理2025/7/291. 問題の内容与えられた極限を計算します。limx→1(x1−x−1logx)(x−1)\lim_{x \to 1} \left( \frac{x}{1-x} - \frac{1}{\log x} \right) (x-1)x→1lim(1−xx−logx1)(x−1)2. 解き方の手順まず、式を整理します。limx→1(xlogx−(1−x)(1−x)logx)(x−1)=limx→1xlogx+x−1(1−x)logx(x−1)=limx→1xlogx+x−1−logx\lim_{x \to 1} \left( \frac{x \log x - (1-x)}{(1-x) \log x} \right) (x-1) = \lim_{x \to 1} \frac{x \log x + x - 1}{(1-x) \log x} (x-1) = \lim_{x \to 1} \frac{x \log x + x - 1}{-\log x}x→1lim((1−x)logxxlogx−(1−x))(x−1)=x→1lim(1−x)logxxlogx+x−1(x−1)=x→1lim−logxxlogx+x−1ここで x=1+hx = 1 + hx=1+h とおくと、x→1x \to 1x→1 のとき h→0h \to 0h→0 なので、limh→0(1+h)log(1+h)+h−log(1+h)\lim_{h \to 0} \frac{(1+h) \log (1+h) + h}{-\log (1+h)}h→0lim−log(1+h)(1+h)log(1+h)+hlog(1+h)\log(1+h)log(1+h) をテイラー展開すると、log(1+h)=h−h22+O(h3)\log (1+h) = h - \frac{h^2}{2} + O(h^3)log(1+h)=h−2h2+O(h3) なので、limh→0(1+h)(h−h22+O(h3))+h−h+h22+O(h3)=limh→0h−h22+h2+O(h3)+h−h+h22+O(h3)=limh→02h+h22+O(h3)−h+h22+O(h3)\lim_{h \to 0} \frac{(1+h)(h - \frac{h^2}{2} + O(h^3)) + h}{-h + \frac{h^2}{2} + O(h^3)} = \lim_{h \to 0} \frac{h - \frac{h^2}{2} + h^2 + O(h^3) + h}{-h + \frac{h^2}{2} + O(h^3)} = \lim_{h \to 0} \frac{2h + \frac{h^2}{2} + O(h^3)}{-h + \frac{h^2}{2} + O(h^3)}h→0lim−h+2h2+O(h3)(1+h)(h−2h2+O(h3))+h=h→0lim−h+2h2+O(h3)h−2h2+h2+O(h3)+h=h→0lim−h+2h2+O(h3)2h+2h2+O(h3)分子と分母を hhh で割ると、limh→02+h2+O(h2)−1+h2+O(h2)=2−1=−2\lim_{h \to 0} \frac{2 + \frac{h}{2} + O(h^2)}{-1 + \frac{h}{2} + O(h^2)} = \frac{2}{-1} = -2h→0lim−1+2h+O(h2)2+2h+O(h2)=−12=−23. 最終的な答え-2