関数の極限の基本的な演習です。
(例題1)次の極限値を求めよ。
(1)\(\displaystyle\lim_{x \to 1}\displaystyle\frac{\sqrt{x}-1}{x^3-1}\)
(2)\(\displaystyle\lim_{x \to -\infty}x(\sqrt{x^2+6x+10}+x+3)\)
(解答)
(1)
\(\displaystyle\lim_{x \to 1}\displaystyle\frac{\sqrt{x}-1}{x^3-1}\)
\(=\displaystyle\lim_{x \to 1}\displaystyle\frac{\sqrt{x}-1}{x^3-1}\cdot\displaystyle\frac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\displaystyle\lim_{x \to 1}\displaystyle\frac{x-1}{(x-1)(x^2+x+1)(\sqrt{x}+1)}\)
(\(x\)を\(1\)に近づけるから、\(1\)そのものになっているわけでないので割れる)
\(=\displaystyle\lim_{x \to 1}\displaystyle\frac{1}{(x^2+x+1)(\sqrt{x}+1)}\)
\(=\displaystyle\frac{1}{3\cdot2}\)
\(=\displaystyle\frac{1}{6}\)
(2)
\(\displaystyle\lim_{x \to -\infty}x(\sqrt{x^2+6x+10}+x+3)\)
\(=\displaystyle\lim_{x \to -\infty}x(\sqrt{x^2+6x+10}+x+3)\cdot\displaystyle\frac{\sqrt{x^2+6x+10}-(x+3)}{\sqrt{x^2+6x+10}-(x+3)}\)
\(=\displaystyle\lim_{x \to -\infty}x\cdot\displaystyle\frac{(x^2+6x+10)-(x+3)^2}{\sqrt{x^2+6x+10}-(x+3)}\)
\(=\displaystyle\lim_{x \to -\infty}\displaystyle\frac{x}{\sqrt{x^2+6x+10}-(x+3)}\)
\(=\displaystyle\lim_{x \to -\infty}\displaystyle\frac{1}{\displaystyle\frac{1}{x}\cdot\sqrt{x^2+6x+10}-(1+\displaystyle\frac{1}{x})}\)
\(=\displaystyle\lim_{x \to -\infty}\displaystyle\frac{1}{\displaystyle\frac{1}{-\sqrt{x^2}}\cdot\sqrt{x^2+6x+10}-(1+\displaystyle\frac{1}{x})}\)
\(=\displaystyle\lim_{x \to -\infty}\displaystyle\frac{1}{-\sqrt{1+\displaystyle\frac{6}{x}+\displaystyle\frac{10}{x^2}}-(1+\displaystyle\frac{1}{x})}\)
\(=\displaystyle\frac{1}{-1-1}\)
\(=-\displaystyle\frac{1}{2}\)
(別解)
\(x=-t\) とすると、\(x \to -\infty\) のとき \(t \to \infty\) だから
\(\displaystyle\lim_{x \to -\infty}x(\sqrt{x^2+6x+10}+x+3)\)
\(=\displaystyle\lim_{t \to \infty}(-t)\{\sqrt{(-t)^2+6(-t)+10}+(-t)+3\}\)
\(=\displaystyle\lim_{t \to \infty}(-t)\{\sqrt{t^2-6t+10}-(t-3)\}\)
\(=\displaystyle\lim_{t \to \infty}(-t)\{\sqrt{t^2-6t+10}-(t-3)\}\cdot\displaystyle\frac{\sqrt{t^2-6t+10}+(t-3)}{\sqrt{t^2-6t+10}+(t-3)}\)
\(=\displaystyle\lim_{t \to \infty}(-t)\cdot\displaystyle\frac{(t^2-6t+10)-(t-3)^2}{\sqrt{t^2-6t+10}+(t-3)}\)
\(=\displaystyle\lim_{t \to \infty}\displaystyle\frac{-t}{\sqrt{t^2-6t+10}+(t-3)}\)
\(=\displaystyle\lim_{t \to \infty}\displaystyle\frac{-1}{\sqrt{1-\displaystyle\frac{6}{t}+\displaystyle\frac{10}{t^2}}+(1-\displaystyle\frac{3}{t})}\)
\(=-\displaystyle\frac{1}{2}\)
(例題2)
(1)\(\displaystyle\lim_{x \to 1-0}\displaystyle\frac{x(x+3)}{x-1}\)、\(\displaystyle\lim_{x \to 1+0}\displaystyle\frac{x(x+3)}{x-1}\) をそれぞれ求めよ。
(2)\(\displaystyle\lim_{x \to \sqrt{2}+0}\displaystyle\frac{x^2-2}{|x^2-3\sqrt{2}x+4|}\) を求めよ。
(解答)
(1)
\(\displaystyle\lim_{x \to 1-0}\displaystyle\frac{x(x+3)}{x-1}=-\infty\) (\(\displaystyle\frac{4}{-0}\)の形)
\(\displaystyle\lim_{x \to 1+0}\displaystyle\frac{x(x+3)}{x-1}=\infty\) (\(\displaystyle\frac{4}{+0}\)の形)
(2)
\(\displaystyle\lim_{x \to \sqrt{2}+0}\displaystyle\frac{x^2-2}{|x^2-3\sqrt{2}x+4|}\)
\(=\displaystyle\lim_{x \to \sqrt{2}+0}\displaystyle\frac{(x-\sqrt{2})(x+\sqrt{2})}{|(x-\sqrt{2})(x-2\sqrt{2})|}\)
\(=\displaystyle\lim_{x \to \sqrt{2}+0}\displaystyle\frac{(x-\sqrt{2})(x+\sqrt{2})}{(x-\sqrt{2})|x-2\sqrt{2}|}\)
\(=\displaystyle\lim_{x \to \sqrt{2}+0}\displaystyle\frac{x+\sqrt{2}}{|x-2\sqrt{2}|}\)
\(=\displaystyle\frac{2\sqrt{2}}{|-\sqrt{2}|}\)
\(=2\)
以上になります。お疲れさまでした。
ここまで見ていただきありがとうございました。
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