関数の変化をとらえる - 関数の極限と微分法関数の極限極限値の計算収束、定数、式の変形

学習環境

新装版数学読本4 (松坂和夫(著)、岩波書店)の第17章(関数の変化をとらえる - 関数の極限と微分法)、17.1(関数の極限)、極限値の計算の問4の解答を求めてみる。

\begin{array}{l} \lim_{x \to 1} (x - 1) \cdot \frac{x^{2} + a x + b}{x - 1} = 0 \cdot 5 = 0 \\ \lim_{x \to 1} (x^{2} + a x + b) = 0 \\ 1 + a + b = 0 \\ b = - a - 1 \end{array}

また、

\begin{array}{l} \lim_{x \to 1} \frac{x^{2} + a x + b}{x - 1} = 5 \\ \lim_{x \to 1} \frac{x^{2} + a x - a - 1}{x - 1} = 5 \\ \lim_{x \to 1} \frac{(x - 1) (x + a + 1)}{x - 1} = 5 \\ \lim_{x \to 1} (x + a + 1) = 5 \\ 1 + a + 1 = 5 \\ a = 3 \end{array}

よって、

a = 3, b = - 4

実際に確認。

\begin{array}{l} \lim_{x \to 1} \frac{x^{2} + 3 x - 4}{x - 1} \\ = \lim_{x \to 1} \frac{(x - 1) (x + 4)}{x - 1} \\ = \lim_{x \to 1} (x + 4) \\ = 5 \end{array}

\begin{array}{l} \lim_{x \to 3} (x^{2} - (3 + b) x + 3 b) \cdot \frac{x^{2} + a x + 6}{x^{2} - (3 + b) x + 3 b} = 0 \cdot \frac{2}{5} = 0 \\ \lim_{x \to 3} (x^{2} + a x + 6) = 0 \\ 9 + 3 a + 6 = 0 \\ a = - 5 \\ \lim_{x \to 3} \frac{x^{2} - 5 x + 6}{x^{2} - (3 + b) x + 3 b} = \frac{2}{5} \\ \lim_{x \to 3} \frac{(x - 3) (x - 2)}{(x - 3) (x - b)} = \frac{2}{5} \\ \lim_{x \to 3} \frac{x - 2}{x - b} = \frac{2}{5} \\ \frac{3 - 2}{3 - b} = \frac{2}{5} \\ 5 = 2 (3 - b) \\ b = \frac{1}{2} \end{array}

\begin{array}{l} \lim_{x \to 2} (x^{3} + a x + b) \cdot \frac{x^{2} - x - 2}{x^{3} + a x + b} = (8 + 2 a + b) \cdot \frac{1}{3} \\ 0 = \frac{1}{3} (8 + 2 a + b) \\ 8 + 2 a + b = 0 \\ b = - 2 a - 8 \\ \lim_{x \to 2} \frac{x^{2} - x - 2}{x^{3} + a x - 2 a - 8} = \frac{1}{3} \\ \lim_{x \to 2} \frac{(x - 2) (x + 1)}{(x - 2) (x^{2} + 2 x + a + 4)} = \frac{1}{3} \\ \lim_{x \to 2} \frac{x + 1}{x^{2} + 2 x + a + 4} = \frac{1}{3} \\ \frac{3}{12 + a} = \frac{1}{3} \\ 9 = 12 + a \\ a = - 3 \\ b = - 2 \end{array}

\begin{array}{l} \lim_{x \to 1} (x - 1) \frac{a \sqrt{x + 1} - b}{x - 1} = 0 \cdot \sqrt{2} = 0 \\ a \sqrt{2} - b = 0 \\ b = a \sqrt{2} \\ \lim_{x \to 1} \frac{a \sqrt{x + 1} - a \sqrt{2}}{x - 1} = \sqrt{2} \\ \lim_{x \to 1} \frac{a (\sqrt{x + 1} - \sqrt{2})}{x - 1} = \sqrt{2} \\ \lim_{x \to 1} \frac{a (x + 1 - 2)}{(x - 1) (\sqrt{x + 1} + \sqrt{2})} = \sqrt{2} \\ \lim_{x \to 1} \frac{a}{\sqrt{x + 1} + \sqrt{2}} = \sqrt{2} \\ \frac{a}{2 \sqrt{2}} = \sqrt{2} \\ a = 4 \\ b = 4 \sqrt{2} \end{array}