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有理式、分母、零、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3の解答を求めてみる。

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

\begin{array}{l} \lim_{t \to 0} \frac{t + t^{4}}{2 t + t^{2}} \\ = \lim_{t \to 0} \frac{1 + t^{3}}{2 + t} \\ = \frac{1}{2} \end{array}

\begin{array}{l} \lim_{x \to - 3} \frac{x^{2} - 9}{x + 3} \\ = \lim_{x \to - 3} \frac{(x + 3) (x - 3)}{x + 3} \\ = \lim_{x \to - 3} (x - 3) \\ = - 6 \end{array}

\begin{array}{l} \lim_{x \to - 6} \frac{x + 6}{x^{2} + 3 x - 18} \\ = \lim_{x \to - 6} \frac{x + 6}{(x + 6) (x - 3)} \\ = \lim_{x \to - 6} \frac{1}{x - 3} \\ = - \frac{1}{9} \end{array}

\begin{array}{l} \lim_{x \to 2} \frac{x^{3} - 8}{x - 2} \\ = \lim_{x \to 2} \frac{(x - 2) (x^{2} + 2 x - 4)}{x - 2} \\ = \lim_{x \to 2} (x^{2} + 2 x - 4) \\ = 4 + 4 - 4 \\ = 4 \end{array}

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

\begin{array}{l} \lim_{x \to 4} \frac{x - 4}{\sqrt{x + 5} - 3} \\ = \lim_{x \to 4} \frac{(x - 4) (\sqrt{x + 5} + 3)}{(\sqrt{x + 5} - 3) (\sqrt{x + 5} + 3)} \\ = \lim_{x \to 4} \frac{(x - 4) (\sqrt{x + 5} + 3)}{x + 5 - 9} \\ = \lim_{x \to 4} \frac{(x - 4) (\sqrt{x + 5} + 3)}{x - 4} \\ = \lim_{x \to 4} (\sqrt{x + 5} + 3) \\ = \sqrt{4 + 5} + 3 \\ = 6 \end{array}

\begin{array}{l} \lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h} \\ = \lim_{h \to 0} \frac{(\sqrt{9 + h} - 3) (\sqrt{9 + h} + 3)}{h (\sqrt{9 + h} + 3)} \\ = \lim_{h \to 0} \frac{9 + h - 9}{h (\sqrt{9 + h} + 3)} \\ = \lim_{h \to 0} \frac{h}{h (\sqrt{9 + h} + 3)} \\ = \lim_{h \to 0} \frac{1}{\sqrt{9 + h} + 3} \\ = \frac{1}{\sqrt{9} + 3} \\ = \frac{1}{6} \end{array}