【微分積分】2-1-1 関数の極限の性質｜問題集

\(=2\)

\(=4\)

\(=-1\)

\(=12\)

\(\displaystyle =\frac{1}{2}\)

\(=-2\)

\(\displaystyle =\lim_{x\to0}\frac{x(x+1)}{2}\)
\(=0\)

\(\displaystyle =\lim_{x\to0}\frac{x+1}{2}\)
\(\displaystyle =\frac{1}{2}\)

\(\displaystyle =\lim_{x\to3}\frac{(x-3)(x+2)}{x-3}\)
\(\displaystyle =\lim_{x\to3}(x+2)\)
\(=5\)

\(\displaystyle =\lim_{x\to4}\frac{(x-4)^2(x+1)^2}{x-4}\)
\(\displaystyle =\lim_{x\to4}(x-4)(x+1)^2\)
\(=0\)

\(\displaystyle =\lim_{x\to1}\frac{(x-1)(x^2+x+1)}{x-1}\)
\(\displaystyle =\lim_{x\to1}(x^2+x+1)\)
\(=3\)

\(\displaystyle =\lim_{x\to2}\frac{(x-1)(x-2)}{(x-2)(x+2)}\)
\(\displaystyle =\lim_{x\to2}\frac{x-1}{x+2}\)
\(\displaystyle =\frac{1}{4}\)

\(\displaystyle =\lim_{x\to1}\frac{(x-1)(2x^3-4x^2-3x-3)}{x-1}\)
\(\displaystyle =\lim_{x\to1}(2x^3-4x^2-3x-3)\)
\(=-8\)

\(\displaystyle =\lim_{x\to\infty}\frac{2+\frac{1}{x^2}}{1+\frac{3}{x}+\frac{2}{x^2}}\)
\(=2\)

\(\displaystyle =\lim_{x\to\infty}\frac{-3x+7+\frac{4}{x}}{2+\frac{5}{x}}\)
\(=-\infty\)

\(\displaystyle =\lim_{x\to9}\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{\sqrt{x}-3}\)
\(\displaystyle =\lim_{x\to9}(\sqrt{x}+3)\)
\(=6\)

\(\displaystyle =\lim_{x\to0}\frac{(\sqrt{x^2+4}-2)(\sqrt{x^2+4}+2)}{x^2(\sqrt{x^2+4}+2)}\)
\(\displaystyle =\lim_{x\to0}\frac{x^2}{x^2(\sqrt{x^2+4}+2)}\)
\(\displaystyle =\lim_{x\to0}\frac{1}{\sqrt{x^2+4}+2}\)
\(\displaystyle =\frac{1}{4}\)

\(\displaystyle =\lim_{x\to0}\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})}\)
\(\displaystyle =\lim_{x\to0}\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}\)
\(\displaystyle =\lim_{x\to0}\frac{2}{\sqrt{1+x}+\sqrt{1-x}}\)
\(=1\)

\(\displaystyle =\lim_{x\to-\infty}x^3\left(1-\frac{4}{x}-\frac{5}{x^2}-\frac{6}{x^3}\right)\)
\(=-\infty\)

\(\displaystyle =\lim_{x\to\infty}\frac{-x+2}{\sqrt{x^2-x+2}+x}\)
\(\displaystyle =\lim_{x\to\infty}\frac{-1+\frac{2}{x}}{\sqrt{1-\frac{1}{x}+\frac{2}{x^2}}+1}\)
\(\displaystyle =-\frac{1}{2}\)

\(t=-x\)とおくと、\(x\to-\infty\)のとき、\(t\to\infty\)
\(\displaystyle =\lim_{t\to\infty}(\sqrt{9t^2-2t}-3t)\)
\(\displaystyle =\lim_{t\to\infty}\frac{-2t}{\sqrt{9t^2-2t}+3t}\)
\(\displaystyle =\lim_{t\to\infty}\frac{-2}{\sqrt{9-\frac{2}{t}}+3}\)
\(\displaystyle =-\frac{1}{3}\)

\(\displaystyle =\lim_{x\to0}\frac{1}{x}\left\{\frac{3-(x+3)}{3(x+3)}\right\}\)
\(\displaystyle =\lim_{x\to0}\frac{1}{x}\left\{\frac{-x}{3(x+3)}\right\}\)
\(\displaystyle =\lim_{x\to0}\frac{-1}{3(x+3)}\)
\(\displaystyle =-\frac{1}{9}\)

\(\displaystyle =\lim_{x\to0}x\left\{\frac{x-(x+1)}{x}\right\}\)
\(=-1\)

\(\displaystyle a-b=\frac{a^3-b^3}{a^2+ab+b^2}\)より、 \(\displaystyle =\lim_{x\to0}\frac{(1+x)-(1-x)}{x\{(1+x)^{\frac{2}{3}}+(1+x)^{\frac{1}{3}}(1-x)^{\frac{1}{3}}+(1-x)^{\frac{2}{3}}\}}\)
\(\displaystyle =\lim_{x\to0}\frac{2}{\{(1+x)^{\frac{2}{3}}+(1+x)^{\frac{1}{3}}(1-x)^{\frac{1}{3}}+(1-x)^{\frac{2}{3}}\}}\)
\(\displaystyle =\frac{2}{3}\)

\(\displaystyle =\lim_{x\to2}x^2\)
\(=4\)