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倍、和、差、積、商

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1の解答を求めてみる。

\begin{array}{l} \lim_{x \to 3} f (x) \\ = \lim_{x \to 3} (2 x - 3) \\ = 6 - 3 \\ = 3 \end{array}

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

\begin{array}{l} \lim_{x \to 3} \frac{1}{3} f (x) \\ = \frac{1}{3} \cdot 3 \\ = 1 \end{array}

\begin{array}{l} \lim_{n \to 3} (- 5) g (x) \\ = - 5 \cdot 6 \\ = - 30 \end{array}

\begin{array}{l} \lim_{x \to 3} (f (x) + g (x)) \\ = \lim_{x \to 3} f (x) + \lim_{x \to 3} g (x) \\ = 3 + 6 \\ = 9 \end{array}

\begin{array}{l} \lim_{x \to 3} (f (x) - g (x)) \\ = \lim_{x \to 3} f (x) - \lim_{x \to 3} g (x) \\ = 3 - 6 \\ = - 3 \end{array}

\begin{array}{l} \lim_{x \to 3} f (x) g (x) \\ = \lim_{x \to 3} f (x) \lim_{x \to 3} g (x) \\ = 3 \cdot 6 \\ = 18 \end{array}

\begin{array}{l} \lim_{x \to 3} \frac{f (x)}{g (x)} \\ = \frac{\lim_{x \to 3} f (x)}{\lim_{x \to 3} g (x)} \\ = \frac{3}{6} \\ = \frac{1}{2} \end{array}