上册 3.1 一元函数的导数 第27题

\section*{解题过程：} （1）由 $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{x}=A$ 知 $\lim _{x \rightarrow 0} f(x)=0$ 。而 $f(x)$ 连续，所以 $f(0)=0, \varphi(0)=0$ ． 当 $x \neq 0$ 时，令 $u=x t$ ，则 $\displaystyle \varphi(x)=\frac{\int_{0}^{x} f(u) \mathrm{d} u}{x}$ ，从而 $\displaystyle \varphi^{\prime}(x)=\frac{x f(x)-\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}$ ． 又因为 $\displaystyle \lim _{x \rightarrow 0} \frac{\varphi(x)-\varphi(0)}{x-0}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}=\lim _{x \rightarrow 0} \frac{f(x)}{2 x}=\frac{A}{2}$ ，即 $\displaystyle \varphi^{\prime}(0)=\frac{A}{2}$ ． 所以 $\displaystyle \varphi^{\prime}(x)=\left\{\begin{array}{l}\frac{x f(x)-\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}, x \neq 0, \\ \frac{A}{2}, x=0 .\end{array}\right.$ 由于 $$ \lim _{x \rightarrow 0} \varphi^{\prime}(x)=\lim _{x \rightarrow 0} \frac{x f(x)-\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}=\lim _{x \rightarrow 0} \frac{f(x)}{x}-\lim _{x \rightarrow 0} \frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}=\frac{A}{2}=\varphi^{\prime}(0) $$ 所以 $\varphi^{\prime}(x)$ 在 $x=0$ 处连续． （2）令 $u=x t$ ，则 $\displaystyle g(x)=\int_{0}^{1} f(x t) \mathrm{d} t=\frac{1}{x} \int_{0}^{x} f(u) \mathrm{d} u,(x \neq 0)$ 。显然，$g(0)=\int_{0}^{1} f(0) \mathrm{d} t=0$ ． $$ \lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{1} f(x t) \mathrm{d} t=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}=\lim _{x \rightarrow 0} \frac{f(x)}{2 x}=\frac{1}{2} \lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\frac{1}{2} f^{\prime}(0)=\frac{5}{2} $$ （3）当 $x=0$ 时，$a \varphi(0)=\int_{0}^{1} \varphi(0) \mathrm{d} t=0$ ，所以 $\varphi(0)=0$ ． 当 $x \neq 0$ 时，令 $u=x t$ ，则 $\displaystyle a \varphi(x)=\int_{0}^{1} \varphi(x t) \mathrm{d} t=\frac{1}{x} \int_{0}^{x} \varphi(u) \mathrm{d} u$ 。于是 $$ a \varphi^{\prime}(x)=-\frac{1}{x^{2}} \int_{0}^{x} \varphi(u) \mathrm{d} u+\frac{1}{x} \varphi(x)=-\frac{a}{x} \varphi(x)+\frac{1}{x} \varphi(x)=(1-a) \frac{1}{x} \varphi(x) $$ 所以，当 $a=1$ 时，$\varphi^{\prime}(x)=0, \varphi(x)=C$ ，从而 $\varphi(x)=0$ ． 当 $a \neq 1$ 且 $a \neq 0$ 时，解方程 $\displaystyle a \varphi^{\prime}(x)=(1-a) \frac{1}{x} \varphi(x)$ ，得 $\displaystyle \varphi(x)=C x^{\frac{1-a}{a}}$ ． （4）因为 $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)-\tan x}{x}=1$ ，所以 $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{x}=2$ 。所以当 $x \neq 0$ 时，令 $u=x t$ ，则 于是 $$ \begin{aligned} & F(x)=\int_{0}^{1} f(t x) \mathrm{d} t=\frac{1}{x} \int_{0}^{x} f(u) \mathrm{d} u \\ & F^{\prime}(x)=\frac{f(x)}{x}-\frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}} \end{aligned} $$ 又 $F(0)=\int_{0}^{1} f(0) \mathrm{d} t=0$ ，所以 $\displaystyle F^{\prime}(0)=\lim _{x \rightarrow 0} \frac{F(x)-F(0)}{x}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}=\lim _{x \rightarrow 0} \frac{f(x)}{2 x}=1$ ． 所以 $\displaystyle F^{\prime}(x)=\left\{\begin{array}{l}\frac{x f(x)-\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}, x \neq 0, \\ 1, x=0 .\end{array}\right.$ 而 $\displaystyle \lim _{x \rightarrow 0} F^{\prime}(x)=\lim _{x \rightarrow 0}\left(\frac{f(x)}{x}-\frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}\right)=\lim _{x \rightarrow 0} \frac{f(x)}{2 x}=1=F^{\prime}(0)$ ，所以 $F^{\prime}(x)$ 在 $x=0$ 连续，从而在 $\mathbf{R}$上连续． （5）因为 $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)-\sin x}{x}=1$ ，所以 $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{x}=2$ ．所以当 $x \neq 0$ 时，令 $u=x t$ ，则 $$ \begin{gathered} F(x)=\int_{0}^{1} f(t x) \mathrm{d} t=\frac{1}{x} \int_{0}^{x} f(u) \mathrm{d} u \\ F^{\prime}(x)=\frac{f(x)}{x}-\frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}} \end{gathered} $$ 于是 又 $F(0)=\int_{0}^{1} f(0) \mathrm{d} t=0$ ，所以 $\displaystyle F^{\prime}(0)=\lim _{x \rightarrow 0} \frac{F(x)-F(0)}{x}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}=\lim _{x \rightarrow 0} \frac{f(x)}{2 x}=1$ ． 所以 $\displaystyle F^{\prime}(x)=\left\{\begin{array}{l}\frac{x f(x)-\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}, x \neq 0, \\ 1, x=0 .\end{array}\right.$ 而 $\displaystyle \lim _{x \rightarrow 0} F^{\prime}(x)=\lim _{x \rightarrow 0}\left(\frac{f(x)}{x}-\frac{\int_{0}^{x} f(u) \mathrm{d} u}{x^{2}}\right)=\lim _{x \rightarrow 0} \frac{f(x)}{2 x}=1=F^{\prime}(0)$ ，所以 $F^{\prime}(x)$ 在 $x=0$ 连续，从而在 $\mathbf{R}$ t．连续． （6）当 $x \neq 0$ 时，令 $u=x^{2} t$ ，则 $\displaystyle f(x)=\int_{0}^{\sin x} g\left(x^{2} t\right) \mathrm{d} t=\frac{1}{x^{2}} \int_{0}^{x^{2} \sin x} g(u) \mathrm{d} u$ ，且 $f(0)=0$ ． $$ \begin{aligned} & f^{\prime}(x)=\frac{\left(x^{2} \sin x\right)^{\prime}}{x^{2}} g\left(x^{2} \sin x\right)-\frac{2}{x^{3}} \int_{0}^{x^{2} \sin x} g(u) \mathrm{d} u=\frac{2 \sin x+x \cos x}{x} g\left(x^{2} \sin x\right)-\frac{2}{x^{3}} \int_{0}^{x^{2} \sin x} g(u) \mathrm{d} u . \\ & f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2} \sin x} g(u) \mathrm{d} u}{x^{3}}=\lim _{x \rightarrow 0} \frac{\left(2 x \sin x+x^{2} \cos x\right) g\left(x^{2} \sin x\right)}{3 x^{2}}=g(0) \end{aligned} $$ 于是 $\displaystyle \quad f^{\prime}(x)=\left\{\begin{array}{l}\frac{2 \sin x+x \cos x}{x} g\left(x^{2} \sin x\right)-\frac{2}{x^{3}} \int_{0}^{x^{2} \sin x} g(u) \mathrm{d} u, x \neq 0, \\ g(0), x=0 .\end{array}\right.$ $$ \begin{aligned} \lim _{x \rightarrow 0} f^{\prime}(x) & =\lim _{x \rightarrow 0}\left(\frac{2 \sin x+x \cos x}{x} g\left(x^{2} \sin x\right)-\frac{2}{x^{3}} \int_{0}^{x^{2} \sin x} g(u) \mathrm{d} u\right) \\ & =3 g(0)-2 \lim _{x \rightarrow 0} \frac{\left(2 x \sin x+x^{2} \cos x\right) g\left(x^{2} \sin x\right)}{3 x^{2}}=3 g(0)-2 g(0)=g(0) \end{aligned} $$ 所以 $f^{\prime}(x)$ 在 $x=0$ 连续，从而 $f^{\prime}(x)$ 在 $(-\infty,+\infty)$ 上连续．