第一章总复习题课

一、填空题

1.\begin{aligned} \lim_{x \to \infty}(\frac{x+2}{x+1})^{3x+1}=[\ e^3\ ] \end{aligned}

解析：\begin{aligned} \lim_{x \to \infty}(\frac{x+2}{x+1})^{3x+1} &=\lim_{x \to \infty}(1+\frac1{x+1})^{3(x+1)-2}\\ &=\lim_{x \to \infty}[(1+\frac1{x+1})^{x+1}]^3(1+\frac1{x+1})^{-2}\\ &=e^3 \end{aligned}

2.\begin{aligned} \lim_{x \to \infty}(\frac{x+3}{x+2})^x=[\ e\ ] \end{aligned}

解析：\begin{aligned} \lim_{x \to \infty}(\frac{x+3}{x+2})^x &=\lim_{x\to\infty}(1+\frac1{x+2})^{x+2-2}\\ &=e \end{aligned}

3.\begin{aligned} \lim_{x \to 0}\frac{1-\sqrt{{\rm cos}\ x}}{x\ln(1+x)} =[\ \frac14\ ] \end{aligned}

4.\begin{aligned} \lim_{x \to 0}\frac{\sqrt{1+x{\rm sin}\ x}-1}{e^{x^2}-1}=[\ \frac12\ ] \end{aligned}

解析：\begin{aligned} \lim_{x \to 0}\frac{\sqrt{1+x{\rm sin}\ x}-1}{e^{x^2}-1} &=\lim_{x\to0}\frac{\frac12x{\rm sin}\ x}{x^2}\\ &=\frac12\lim_{x\to0}\frac{x\cdot x}{x^2}\\ &=\frac12 \end{aligned}

5.\begin{aligned} \lim_{x\to \infty}x[\ln(x+1)-\ln x] =[\ 1\ ] \end{aligned}

解析：\begin{aligned} \lim_{x\to \infty}x[\ln(x+1)-\ln x] &=\lim_{x\to\infty}\ln (1+\frac1x)^x\\ &=\ln[\lim_{x\to \infty}(1+\frac1x)^x]\\ &=\ln e=1 \end{aligned}

6.\begin{aligned} \lim_{x\to0}e^{\frac{\normalsize\tan x}{\normalsize x}}=[\ e\ ] \end{aligned}

解析：\begin{aligned} \lim_{x\to0}e^{\frac{\normalsize\tan x}{\normalsize x}} &=e^{\normalsize\lim_{x\to0}\frac{\normalsize\tan x}{\normalsize x}}\\ &=e \end{aligned}

二、选择题

7.\begin{aligned} \lim_{x \to \infty}\frac{x^2+x-2}{x-\cos x}= \end{aligned}[ B ]

A.-1

B.\infty

C.1

D.2

解析：\begin{aligned} \lim_{x \to \infty}\frac{x^2+x-2}{x-\cos x}&=\lim_{x\to\infty}\frac{2x+1}{1-(-\sin x)}\\ &=\lim_{x\to\infty}\frac{2x+1}{\sin x+1}\\ &=\infty \end{aligned}

8.\begin{aligned} \lim_{x\to\infty}(\frac{x^2}{x+1}-\frac{x^2}{x-1})= \end{aligned}[ D ]

A.1

B. 0

C.-1

D.-2

解析：\begin{aligned} \lim_{x\to\infty}(\frac{x^2}{x+1}-\frac{x^2}{x-1}) &=\lim_{x\to\infty}\frac{-2x^2}{x^2-1}\\ &=\lim_{x\to\infty}\frac{-2}{1-\frac{1}{x^2}}\\ &=-2 \end{aligned}

9.已知\begin{aligned} \lim_{x\to1}\frac{x^2-3x+c}{x-1}=-1 \end{aligned}，则c的值为[ C ].

A.-1

B.1

C.2

D.3

解析：\begin{aligned} &\lim_{x\to1}\frac{x^2-3x+c}{x-1}=-1\\ &x^2-3x+c=(x-1)(x+a)\\ &=x^2+(a-1)x-a\\ &a-1=-3,a=-2\\ &c=-a=2 \end{aligned}

10.已知极限\begin{aligned} \lim_{x\to\infty}(\frac{x^2+2}{x}+ax)=0 \end{aligned}，则常数a等于[ A ].

A.-1

B. 0

C.1

D.2

解析：\begin{aligned} \lim_{x\to\infty}(\frac{x^2+2}{x}+ax) &=\lim_{x\to \infty}\left(\frac{(a+1)x^2+2}{x}\right)\\ &=0\\ a+1=0,\therefore a=-1 \end{aligned}

三、解答题

11.设函数\begin{aligned} f(x)=\frac{x^2-1}{x^2-3x+2} \end{aligned}，求f(x)的间断点，并判断间断点类型。

解答：由于f(x)在x=1,x=2点无定义，因此x=1,x=2是f(x)的间断点，

\begin{aligned} \lim_{x\to1}f(x)&=\lim_{x\to1}\frac{x^2-1}{x^2-3x+2}\\ &=\lim_{x\to1}\frac{(x-1)(x+1)}{(x-1)(x-2)}\\ &=-2 \end{aligned}，故x=1是f(x)的可去间断点；

\begin{aligned} \lim_{x\to2}f(x)&=\lim_{x\to2}\frac{x^2-1}{x^2-3x+2}\\ &=\lim_{x\to2}\frac{(x-1)(x+1)}{(x-1)(x-2)}\\ &=\infty \end{aligned}，故x=2是f(x)的无穷间断点。

12.设函数\begin{aligned} f(x)=\frac1{1-{\rm e}^{\frac{x}{1-x}}} \end{aligned}，求f(x)的间断点。

解答：由于f(x)在x=0,x=1点无定义，因此x=0,x=1是f(x)的间断点，

\begin{aligned} \lim_{x\to0}f(x)=\lim_{x\to0}\frac1{1-{\rm e}^{\frac x{1-x}}}=\infty \end{aligned}，故x=0是f(x)的无穷间断点；

\begin{aligned} \lim_{x\to{1^{-}}}f(x)=\lim_{x\to{1^{-}}}\frac 1{1-{\rm e}^{\frac x{1-x}}}=0 \end{aligned}；\begin{aligned} \lim_{x\to{1^{+}}}f(x)=\lim_{x\to{1^{+}}}\frac 1{1-{\rm e}^{\frac x{1-x}}}=1. \end{aligned}

故x=1是f(x)的跳跃间断点。

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