16.证明:\(\lim\limits_{x\rightarrow0}f(x)\)与\(\lim\limits_{x\rightarrow0}f(x^3)\)有一个存在时,另一个也存在,而且两者相等;问是否\(\lim\limits_{x\rightarrow0}f(x)\)与\(\lim\limits_{x\rightarrow0}f(x^2)\)一定同时存在.
证明
若\(\lim\limits_{x\rightarrow0}f(x)=A\),则\(\forall\varepsilon>0,\exists\delta>0\),当\(|x|<\delta,|f(x)-A|<\varepsilon\).
而当\(|x|<\sqrt[3]{\delta},|f(x^3)-A|<\varepsilon\),即\(\lim\limits_{x\rightarrow0}f(x^3)=A\);
若\(\lim\limits_{x\rightarrow0}f(x^3)=A\),则\(\forall\varepsilon>0,\exists\delta>0\),当\(|x|<\delta,|f(x^3)-A|<\varepsilon\).
而当\(|x|<\delta^3,|f(x)-A|<\varepsilon\),即\(\lim\limits_{x\rightarrow0}f(x)=A\).
不一定,令\(f(x)=\lfloor x\rfloor\).
17.指出下列函数的间断点,并说明属于哪一种类型间断点:
(1)\(f(x)=\text{sgn}(\sin x+\frac{1}{2})\);
\(2k\pi-\frac{\pi}{6},2k\pi+\frac{7\pi}{6},k\in\mathbb{Z}\).跳跃间断点.
(2)\(f(x)= \begin{cases} \frac{x}{(1+x)^2},&x\not=-1,\\ 0,&x=-1; \end{cases}\)
\(-1\).第二类间断点.
(3)\(f(x)= \begin{cases} x,&|x|\leqslant1,\\ \ln|x+1|,&|x|>1; \end{cases}\)
\(-1\),第二类间断点.\(1\),跳跃间断点.
(4)\(f(x)= \begin{cases} \cos^2\frac{1}{x},&x\not=0,\\ 1,&x=0. \end{cases}\)
\(0\).第二类间断点.
18.给出下列函数在\(x=0\)的函数值,使其在该点连续:
(1)\(f(x)=\frac{\sqrt[3]{1+x}-1}{\sqrt{1+x}-1}\);
解 \(\frac{2}{3},\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+x}-1}{\sqrt{1+x}-1}=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}+1}{\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}+1}=\frac{2}{3}\)
(2)\(f(x)=\sin x\sin\frac{1}{x}\).
解 \(0,\lim\limits_{x\rightarrow0}|\sin x\sin\frac{1}{x}|\leqslant\lim\limits_{x\rightarrow0}|\sin x|=0,\lim\limits_{x\rightarrow0}\sin x\sin\frac{1}{x}=0\).
19.适当选取\(\alpha\),使函数\(f(x)= \begin{cases} \text{e}^x,&x<0,\\ \alpha+x,&x\geqslant0 \end{cases}\)在\((-\infty,+\infty)\)上连续.
解 \(\alpha=1\).
20.设函数\(f(x)\)在\(x=x_0\)处连续,函数\(g(x)\)在\(x=x_0\)处不连续,问\(f(x)+g(x)\)和\(f(x)g(x)\)是否在\(x=x_0\)处一定不连续.
解 一定;不一定,如\(f(x)\equiv0\).
