【高校数学Ⅲ】2-2-1 関数の極限｜問題集

\(=-1^2+2\)
\(=1\)

\(\displaystyle =2・1^2-3・1-1\)
\(=-2\)

\(\displaystyle =(-2-3)(-2+2)\)
\(=0\)

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

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

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

\(\displaystyle =\lim_{x\to4}\frac{(x-4)(\sqrt{x}+2)}{x-4}\)
\(\displaystyle =\lim_{x\to4}(\sqrt{x}+2)\)
\(=4\)

\(\displaystyle \lim_{x\to3}(x-3)=0\)より、
\(\displaystyle \lim_{x\to3}(a\sqrt{x+1}+b)=0\)
これを解くと、\(b=-2a\)
\(\displaystyle \lim_{x\to3}\frac{a\sqrt{x+1}-2a}{x-3}\)
\(\displaystyle =\lim_{x\to3}\frac{a(x+1-4)}{(x-3)(\sqrt{x+1}+2)}\)
\(\displaystyle =\lim_{x\to3}\frac{a}{\sqrt{x+1}+2}\)
\(\displaystyle =\frac{a}{4}=2\)
よって、
\(a=8,b=-16\)

\(\displaystyle \lim_{x\to1}(x-1)=0\)より、
\(\displaystyle \lim_{x\to1}(a\sqrt{x+8}+b)=0\)
これを解くと、\(b=-3a\)
\(\displaystyle \lim_{x\to1}\frac{a\sqrt{x+8}-3a}{x-1}\)
\(\displaystyle =\lim_{x\to1}\frac{a(x+8-9)}{(x-1)(\sqrt{x+8}+3)}\)
\(\displaystyle =\lim_{x\to1}\frac{a}{\sqrt{x+8}+3}\)
\(\displaystyle =\frac{a}{6}=-1\)
よって、
\(a=-6,b=18\)

\(=0\)

\(=0\)

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

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

\(\displaystyle =\lim_{x\to-\infty}\frac{1+\frac{5}{x}+\frac{7}{x^3}}{1+\frac{3}{x^3}}\)
\(=1\)

\(\displaystyle =\lim_{x\to\infty}\frac{2-\frac{1}{x}}{4+\frac{3}{x}}\)
\(\displaystyle =\frac{1}{2}\)

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

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

\(\displaystyle =\lim_{x\to-\infty}\frac{\frac{3}{x^2}-\frac{2}{x}}{1-\frac{4}{x}+\frac{1}{x^2}}\)
\(=0\)

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

\(a=-x\)とおくと、
\(\displaystyle \lim_{a\to\infty}(\sqrt{4a^2-2a}-2a)\)
\(\displaystyle =\lim_{a\to\infty}\frac{(4a^2-2a)-4a^2}{\sqrt{4a^2-2a}+2a)}\)
\(\displaystyle =\lim_{a\to\infty}\frac{-2a}{\sqrt{4a^2-2a}+2a}\)
\(\displaystyle =\lim_{a\to\infty}\frac{-2}{\sqrt{4-\frac{2}{a}}+2}\)
\(\displaystyle =-\frac{1}{2}\)

\(=-\infty\)

\(=1\)

\(=-1\)

\(x>3\)のとき、
\(\displaystyle \lim_{x\to3+0}\frac{x-3}{x(x-3)}\)
\(\displaystyle =\lim_{x\to3+0}\frac{1}{x}\)
\(\displaystyle =\frac{1}{3}\)

\(x>1\)のとき、
\(\displaystyle \lim_{x\to1+0}\frac{(x+1)(x-1)}{x-1}\)
\(\displaystyle =\lim_{x\to1+0}(x+1)\)
\(\displaystyle =2\)

\(x<1\)のとき、
\(\displaystyle \lim_{x\to1-0}\frac{(x+1)(x-1)}{-(x-1)}\)
\(\displaystyle =\lim_{x\to1-0}-(x+1)\)
\(\displaystyle =-2\)

\(=\infty\)

\(=-\infty\)

\(\displaystyle \lim_{x\to2+0}\frac{1}{(x-2)^2}=\infty\)
\(\displaystyle \lim_{x\to2-0}\frac{1}{(x-2)^2}=\infty\)

\(\displaystyle \lim_{x\to2+0}\frac{1}{(x-2)^2}=\lim_{x\to2-0}\frac{1}{(x-2)^2}\)より、
\(\displaystyle \lim_{x\to2}\frac{1}{(x+1)^2}=\infty\)

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

\(\displaystyle \lim_{x\to-1+0}-\frac{1}{(x+1)^2}=\lim_{x\to-1-0}-\frac{1}{(x+1)^2}\)より、
\(\displaystyle \lim_{x\to-1}-\frac{1}{(x+1)^2}=-\infty\)