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累乗、累乗根

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2の解答を求めてみる。

\begin{array}{l} \lim_{x \to - 2} (- 4 + 3 x) \\ = - 4 - 6 \\ = - 10 \end{array}

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

\begin{array}{l} \lim_{x \to 2} (x^{3} - 3 x^{2} + x + 2) \\ = 8 - 12 + 2 + 2 \\ = 0 \end{array}

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

\begin{array}{l} \lim_{y \to - 1} \frac{y^{2} + y - 3}{2 - y} \\ = \frac{1 - 1 - 3}{3} \\ = - 1 \end{array}

\begin{array}{l} \lim_{t \to - 3} \frac{t^{4} + t^{2} + 6}{t^{2} + t} \\ = \frac{81 + 9 + 6}{9 - 3} \\ = \frac{96}{6} \\ = 16 \end{array}

\begin{array}{l} \lim_{x \to 0} \sqrt{9 - x} \\ = \sqrt{9} \\ = 3 \end{array}

\begin{array}{l} \lim_{x \to - 4} \frac{10}{\sqrt{x^{2} + 2 x + 1}} \\ = \frac{10}{\sqrt{16 - 8 + 1}} \\ = \frac{10}{\sqrt{9}} \\ = \frac{10}{3} \end{array}