【微分積分】2-1-2 右極限と左極限|問題集
1.次の関数の片側極限値を求めなさい。
(1)\(\displaystyle \lim_{x\to+0}\frac{1}{x}\)
\(=\infty\)
(2)\(\displaystyle \lim_{x\to-0}\frac{1}{x}\)
\(=-\infty\)
(3)\(\displaystyle \lim_{x\to+0}\frac{1}{x^2}\)
\(=\infty\)
(4)\(\displaystyle \lim_{x\to-0}\frac{1}{x^2}\)
\(=\infty\)
(5)\(\displaystyle \lim_{x\to1+0}[x]\)
\(=1\)
(6)\(\displaystyle \lim_{x\to1-0}[x]\)
\(=0\)
(7)\(\displaystyle \lim_{x\to2-}\frac{x-2}{|x-2|}\)
\(\displaystyle =\lim_{x\to2-}\frac{x-2}{-(x-2)}\)
\(=-1\)
\(=-1\)
(8)\(\displaystyle \lim_{x\to2+}\frac{x-2}{|x-2|}\)
\(\displaystyle =\lim_{x\to2+}\frac{x-2}{x-2}\)
\(=1\)
\(=1\)
(9)\(\displaystyle \lim_{x\to1-}\sqrt{|x|-x}\)
\(=0\)
(10)\(\displaystyle \lim_{x\to1+}\sqrt{|x|-x}\)
\(=0\)
(11)\(\displaystyle \lim_{x\to1-}\frac{\sqrt{x}-1}{x-1}\)
\(\displaystyle =\lim_{x\to1-}\frac{x-1}{(x-1)(\sqrt{x}+1)}\)
\(\displaystyle =\lim_{x\to1-}\frac{1}{\sqrt{x}+1}\)
\(\displaystyle =\frac{1}{2}\)
\(\displaystyle =\lim_{x\to1-}\frac{1}{\sqrt{x}+1}\)
\(\displaystyle =\frac{1}{2}\)
(12)\(\displaystyle \lim_{x\to0+}\frac{|x|}{\sqrt{a+x}-\sqrt{a-x}}\)
\(\displaystyle =\lim_{x\to0+}\frac{x(\sqrt{a+x}+\sqrt{a-x})}{(\sqrt{a+x}-\sqrt{a-x})(\sqrt{a+x}+\sqrt{a-x})}\)
\(\displaystyle =\lim_{x\to0+}\frac{x(\sqrt{a+x}+\sqrt{a-x})}{a+x-(a-x)}\)
\(\displaystyle =\sqrt{a}\)
\(\displaystyle =\lim_{x\to0+}\frac{x(\sqrt{a+x}+\sqrt{a-x})}{a+x-(a-x)}\)
\(\displaystyle =\sqrt{a}\)
(13)\(\displaystyle \lim_{x\to0-}\frac{x}{\sqrt{1-\cos x}}\)
\(\displaystyle =\lim_{x\to0-}\frac{x\sqrt{1+\cos x}}{\sqrt{1-\cos^2x}}\)
\(\displaystyle =\lim_{x\to0-}\frac{x\sqrt{1+\cos x}}{\sqrt{\sin^2x}}\)
\(\displaystyle =\lim_{x\to0-}\frac{x\sqrt{1+\cos x}}{|\sin x|}\)
\(\displaystyle =\lim_{x\to0-}\frac{x\sqrt{1+\cos x}}{-\sin x}\)
\(=-\sqrt{2}\)
\(\displaystyle =\lim_{x\to0-}\frac{x\sqrt{1+\cos x}}{\sqrt{\sin^2x}}\)
\(\displaystyle =\lim_{x\to0-}\frac{x\sqrt{1+\cos x}}{|\sin x|}\)
\(\displaystyle =\lim_{x\to0-}\frac{x\sqrt{1+\cos x}}{-\sin x}\)
\(=-\sqrt{2}\)
次の学習に進もう!