문제 풀이

2 극한

1.1

$$\begin{array}{rl}& \forall x\ne 0\\ & -1\le \cos \left(\frac{1}{x}\right)\le 1\end{array}$$

$x^4+x^2$이 항상 양수이므로

$$\begin{array}{rl}& -(x^4+x^2)\le \cos \left(\frac{1}{x}\right)\le (x^4+x^2)\\ & \therefore \underset{x\to 0}{\lim }x\sin x=0\end{array}$$

1.2

$$\begin{array}{rl}& \forall x\ne 0\\ & 0\le |x\sin \left(\frac{1}{x^2}\right)|=|x||\sin \left(\frac{1}{x^2}\right)|\le |x|\\ & \underset{x\to 0}{\lim }x=0\\ & \therefore \underset{x\to 0}{\lim }x\sin \left(\frac{1}{x^2}\right)=0\end{array}$$

1.3

1.3.1

자연상수 정의

$$\begin{array}{rl}e& =\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n\\ & =\underset{n\to 0}{\lim }{\left(1+n\right)}^{1/n}\end{array}$$

$$\begin{array}{rl}\underset{n\to \infty }{\lim }{\left(\frac{n-2}{n}\right)}^{3n}& =\underset{n\to \infty }{\lim }{\left(1-\frac{2}{n}\right)}^{3n}\\ & =\underset{n\to \infty }{\lim }{\left(1-\frac{2}{n}\right)}^{-n/2\times -6}\\ & =e^{-6}\end{array}$$

1.3.2

$$\begin{array}{rl}f(x{)}^{g(x)}& =e^{\ln f(x{)}^{g(x)}}(\because f(x{)}^{g(x)}>0)\\ & =e^{g(x)\ln f(x)}(\because f(x)>0)\\ f(x)>0& ⟹f(x{)}^{g(x)}=e^{g(x)\ln f(x)}\end{array}$$

$$\begin{array}{rl}\underset{n\to \infty }{\lim }{\left(\frac{n-2}{n}\right)}^{3n}& =\underset{n\to \infty }{\lim }{e}^{3n\ln \left(\frac{n-2}{n}\right)}\\ & =e^{{\lim }_{n\to \infty }3n\ln \left(\frac{n-2}{n}\right)}\\ & \\ \underset{n\to \infty }{\lim }n\ln \left(\frac{n-2}{n}\right)& =\underset{n\to \infty }{\lim }\frac{\ln (n-2)-\ln n}{\frac{1}{n}}\\ & =\underset{n\to \infty }{\lim }\frac{\left(\frac{1}{n-2}-\frac{1}{n}\right)}{-\frac{1}{n^2}}\\ & =\underset{n\to \infty }{\lim }\frac{\left(\frac{2}{n(n-2)}\right)}{-\frac{1}{n^2}}\\ & =\underset{n\to \infty }{\lim }\frac{-2n}{n-2}\\ & =2\end{array}$$

1.4

$$\begin{array}{r}\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1=\underset{\theta \to 0}{\lim }\frac{\theta }{\sin \theta }\\ \underset{\theta \to 0}{\lim }\frac{\tan \theta }{\theta }=1=\underset{\theta \to 0}{\lim }\frac{\theta }{\tan \theta }\end{array}$$

$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{x{\sin }^{2}x}{{\tan }^{3}x}& =\underset{x\to 0}{\lim }\frac{x^3}{{\tan }^{3}x}\times \frac{\sin x^2}{x^2}\\ & =1\end{array}$$

1.5

$$f(1)=1⟹\underset{x\to 1}{\lim }\ln (\forall f(x))=\ln (\underset{x\to 1}{\lim }\forall f(x))$$

연속이면 극한값과 함수값이 같으므로

$$\underset{x\to 1}{\lim }\ln (f(x))=0$$

$$\begin{array}{r}\theta ={\tan }^{-1}x\\ \tan \theta =x\end{array}$$

이때 역함수가 존재하기 위해서는 $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$를 만족해야한다.

$$\therefore \underset{n\to \infty }{\lim }{\tan }^{-1}(x)=\frac{\pi }{2}$$

$$\begin{array}{rl}\underset{x\to 1}{\lim }(\sin (x){)}^{-1}& =\frac{\pi }{2}\\ \underset{x\to 1}{\lim }(\sin (x){)}^{-1}& =\underset{x\to 1}{\lim }\frac{1}{\sin x}\end{array}$$

${\tan }^{-1}$의 정의역이 $(-\infty ,\infty )$이므로 성립

1.6

$$\begin{array}{rl}\underset{\theta \to 0}{\lim }\frac{\tan \theta -\sin \theta }{{\theta }^3}& =\underset{\theta \to 0}{\lim }\frac{\sin \theta (\sec \theta -1)}{{\theta }^3}\\ & =\underset{\theta \to 0}{\lim }\frac{\sin \theta ({\sec }^2\theta -1)}{{\theta }^3(\sec \theta +1)}\\ & =\underset{\theta \to 0}{\lim }\frac{\sin \theta {\tan }^2\theta }{{\theta }^3(\sec \theta +1)}\\ & =\underset{n\to \infty }{\lim }\frac{\sin \theta }{\theta }\times {\left(\frac{\tan \theta }{\theta }\right)}^2\times \frac{1}{\sec \theta +1}\\ & =\underset{n\to \infty }{\lim }\frac{1}{\sec \theta +1}\\ & =\frac{1}{2}\end{array}$$

1.7

$$\begin{array}{rl}& \\ \underset{x\to 0}{\lim }2x\cot 3x& =\underset{x\to 0}{\lim }\frac{2x}{\tan 3x}\\ & =\underset{x\to 0}{\lim }\frac{3x}{\tan 3x}\times \frac{2}{3}\\ & =\frac{2}{3}\end{array}$$ $$\begin{array}{rl}\underset{x\to 5}{\lim }\frac{4\sin (x-5)}{3x^2-18x+15}& =\underset{x\to 5}{\lim }\frac{4\sin (x-5)}{3(x^2-6x+5)}\\ & =\underset{x\to 5}{\lim }\frac{4\sin (x-5)}{3(x-1)(x-5)}\\ & =\underset{x\to 5}{\lim }\frac{4}{3(x-1)}\\ & =\frac{1}{3}\end{array}$$ $$\therefore \underset{x\to 0}{\lim }2x\cot 3x+\underset{x\to 5}{\lim }\frac{4\sin (x-5)}{3x^2-18x+15}=1$$

1.8

$$\underset{x\to -2}{\lim }\frac{1-\sqrt{x+a}}{x+2}=b$$

위 식이 b로 수렴하기 위해서는 분자가 0이 되어야 한다.

$$\therefore a=3$$ $$\begin{array}{rl}\underset{x\to -2}{\lim }\frac{1-\sqrt{x+3}}{x+2}& =\underset{x\to -2}{\lim }\frac{(1-\sqrt{x+3})(1+\sqrt{x+3})}{(x+2)(1+\sqrt{x+3})}\\ & =\underset{x\to -2}{\lim }\frac{-(x+2)}{(x+2)(1+\sqrt{x+3})}\\ & =\underset{x\to -2}{\lim }\frac{-1}{(1+\sqrt{x+3})}\\ & =-\frac{1}{2}\end{array}$$

1.9

$$\begin{array}{rl}\underset{x\to \infty }{\lim }\frac{x^{-\frac{5}{4}}}{\sin \left(\frac{1}{x}\right)}& =\underset{x\to \infty }{\lim }\frac{x^{-1}{x}^{-\frac{1}{4}}}{\sin \left(\frac{1}{x}\right)}\\ & =\underset{x\to \infty }{\lim }1\times x^{-\frac{1}{4}}\\ & =0\end{array}$$

1.10

$$\begin{array}{r}\underset{x\to 4^{+}}{\lim }\frac{[x{]}^2-16}{x-4}=\underset{x\to 4^{+}}{\lim }\frac{4^2-16}{x-4}=0\end{array}$$

1.11

$$f(x)=\{\begin{array}{l}{c}^2{x}^2+2x,x<2\\ x^3-c^{2}x,x\ge 2\end{array}$$

각각의 조각함수는 이미 연속이므로 2에서 좌극한과 우극한을 일치시키면 된다.

$$\begin{array}{rl}4c^2+4& =8-2c^2\\ c& =\pm \sqrt{\frac{2}{3}}\\ \therefore |c_1-c_2|=2\sqrt{\frac{2}{3}}& \end{array}$$

1.12

$f(x)$가 연속함수이므로

$$\begin{array}{rl}\underset{x\to 0}{\lim }f(x)& =\underset{x\to 0}{\lim }\frac{\sqrt{x^2+16}-4}{x^2}+\frac{\sin x}{3x}=\alpha \\ & =\underset{x\to 0}{\lim }\frac{x^2}{x^2(\sqrt{x^2+16}+4)}+\frac{1}{3}\frac{\sin x}{x}\\ & =\underset{x\to 0}{\lim }\frac{1}{\sqrt{x^2+16}+4}+\frac{1}{3}\\ & =\frac{11}{24}\\ & \therefore \alpha =\frac{11}{24}\end{array}$$

1.13

$$\begin{array}{rl}\underset{x\to 0}{\lim }f(x)& =\underset{x\to 0}{\lim }\left(1-x\sin \left(\frac{1}{e^{4x}}\right)\right)\\ & =1\\ (\because 0& \le |x\sin \left(\frac{1}{e^{4x}}\right)|=|x||\sin \left(\frac{1}{e^{4x}}\right)|\le |x|)\end{array}$$

1.14

우선 $f(x)$가 불연속인 $x$ 좌표를 구해보면 $x=1\text{or}-1$이다. 이를 $g(x)$에 대입하여 불연속인지를 확인해보면

$$\begin{array}{rl}\text{case1)}x=1& \\ \underset{x\to 1^{+}}{\lim }f(x)& =-1\\ \underset{x\to 1^{-}}{\lim }f(x)& =1\\ \underset{x\to 1^{+}}{\lim }g(x)& =\underset{x\to 1^{+}}{\lim }f(f(x))=\underset{x\to 1}{\lim }-|-|x||=-1\\ \underset{x\to 1^{-}}{\lim }g(x)& =\underset{x\to 1^{-}}{\lim }f(f(x))=\underset{x\to 1}{\lim }{x}^2=1\\ & \\ \text{case2)}x=-1& \\ \underset{x\to -1^{+}}{\lim }f(x)& =1\\ \underset{x\to -1^{-}}{\lim }f(x)& =-1\\ \underset{x\to -1^{+}}{\lim }g(x)& =\underset{x\to 1^{+}}{\lim }{f}^2(x)=1\\ \underset{x\to 1^{-}}{\lim }g(x)& =\underset{x\to 1^{-}}{\lim }{f}^2(x)=1\end{array}$$ $$\therefore x=1$$

3 미분

2.1

$$\begin{array}{rl}\underset{n\to \infty }{\lim }n{f}^{-1}\left(\frac{1}{n}\right)& =\underset{t\to 0^{+}}{\lim }\frac{1}{t}{f}^{-1}(t)\left(\because t=\frac{1}{n}\right)\\ & =\underset{t\to 0}{\lim }\frac{f^{-1}(t)-f^{-1}(0)}{t-0}\\ & =\underset{t\to 0}{\lim }(f^{-1}{)}^{\prime }(t)\\ \underset{t\to 0}{\lim }(f^{-1}{)}^{\prime }(x)& =\underset{x\to 0}{\lim }\frac{1}{f^{\prime }(x)}\\ & =\underset{x\to 0}{\lim }\frac{1}{6x^2+3}\\ & =\frac{1}{3}\end{array}$$

2.2

$a$에서 접점을 가짐을 통해 다음을 알 수 있다.

$$\begin{array}{rl}& \\ f(a)=g(a)⟹e^{2a}& =k\sqrt{a}\\ f^{\prime }(a)=g^{\prime }(a)⟹2e^{2a}& =\frac{k}{2\sqrt{a}}\\ & \\ \therefore a=\frac{1}{4},k=2\sqrt{e}& \\ ka=\frac{\sqrt{e}}{2}& \end{array}$$

2.3

라이프니츠 공식

$$\begin{array}{r}(f\cdot g{)}^n(x)=\sum _{k=0}^n{}_n{C}_k{f}^{(n-k)}(x)g^{(k)}(x)\end{array}$$

$$\begin{array}{rl}& (f\cdot g{)}^{(7)}(x)=f^{(7)}(x)g(x)+{}_7{C}_1{f}^{(6)}(x)g^{(1)}(x)+{}_7{C}_2{f}^{(5)}(x)g^{(2)}(x)+\cdots +f(x)g^{(7)}(x)\\ & a_1+a_2+a_3+\cdots +a_6{=}_7{C}_1{+}_7{C}_2{+}_7{C}_3+{\cdots }_7{C}_6\end{array}$$

이항정리

$$\begin{array}{r}(a+b{)}^n=\sum _{k=0}^n{}_n{C}_k{a}^{n-k}{b}^k\end{array}$$

이항정리에서 $a=b=1$인 경우를 생각해보면 $a_1+a_2+a_3+\cdots +a_6=(1+1{)}^7-2$ 따라서 평균은

$$\begin{array}{rl}\frac{a_1+a_2+a_3+\cdots +a_6}{6}& =\frac{(1+1{)}^7-2}{6}\\ & =\frac{126}{6}\\ & =21\end{array}$$

2.4

$x\to 1$에서 수렴하고, 분모가 모두 0이 되므로 분자도 0이 되어야 한다.

\begin{align} \therefore f(1)=3, g(1)=-2 \\ \end{align}$$

\begin{align} \lim_{ x \to 1 }\frac{(f(x)-3)}{x-1}&=\lim_{ x \to 1 }\frac{(f(x)-f(1))}{x-1} \ &=f’(1) \ \lim_{ x \to 1 }\frac{(g(x)+2)}{x-1}&=\lim_{ x \to 1 }\frac{(g(x)-g(1))}{x-1} \ &=g’(1) \end{align}

\begin{align} \therefore f’(1)=1, g’(1)=2 \end{align}

\begin{align} h’(x)&=f’(x)g(x)+f(x)g’(x) \ &=1\times -2 + 3\times 2 \ =4 \end{align}

### 2.5 미분 가능하다 $\implies$ 연속 1) 극한값이 함수값과 같다.

\begin{align} f(1)&=a=a+b+4 \ b&=-4 \end{align}

2) $x=1$에서 미분 가능하다.

\begin{align} f’(x)&=a=2ax+b \ f’(1)&=a=2a+b \ \therefore &=-b=4 \end{align}

\therefore a-b=8

### 2.6 1. $x\to0$에서 극한값이 존재하지 않으므로 불연속이다. 2. 연속

\begin{align} 0\leq |x\sin \frac{1}{x}|\leq|x| \ \therefore \lim_{ x \to 0 } |x \sin \frac{1}{x}|=0 \end{align}

$$미분가능성$$

\begin{align} f’(0)&=\lim_{ x \to 0 } \frac{f(x)-f(0)}{x-0} \ &=\lim_{ x \to 0 } \sin \frac{1}{x} \end{align}

$$연속이지만극한값이존재하지않으므로미분이불가능하다.3.1과마찬가지로연속이고다음과같이미분가능함을보일수있다.$$

\begin{align} \lim_{ x \to 0 } f’(x)&=\lim_{ x \to 0 }\frac{f(x)-f(0)}{x-0} \ &=\lim_{ x \to 0 }x\sin x \ &=0 \end{align}

>$$ >\begin{align} >&\alpha,\beta>0\\ >f(x)=\begin{cases} >x^{\alpha}\sin \frac{1}{x^{\beta}}&(x\neq 0) \\ >0&(x=0) >\end{cases} >\end{align} >$$ >$\alpha>1$이면 $x=0$에서 미분가능하다. ### 2.7 라이프니츠 규칙에 의하여

\begin{align} (f(x)g(x))^{(8)}&=\sum_{k=0}^{8} {}{8}C{k}f^{(8-k)}(x)g^{(k)}(x) \ &={}{8}C{0}f^{(8)}(x)g(x)+{}{8}C{1}f^{(7)}(x)g^{(1)}(x)+\cdots+{}{8}C{8}f(x)g^{(8)}(x) \ &={}{8}C{0}(-2)^{8}(-3)^{0}+{}{8}C{1}(-2)^{7}(-3)+{}{8}C{8}(-2)^{0}(-3)^{8} \ \end{align}

\begin{align} [(-2)+(-3)]^{8}=\sum_{k=0}^{8} {}{8}C{k} (-2)^{8-k}(-3)^{k} \end{align}

\begin{align} \therefore 5^{8} \end{align}

### 2.8

\begin{align} &f(0+0)=2f(0) \ &f(0)=0 \end{align}

\begin{align} \lim_{ h \to 0 } \frac{f(h)-f(0)}{h} = \lim_{ h \to 0 } \frac{f(h)}{h}=1 \end{align}

\begin{align} f’(x)&=\lim_{ h \to 0 } \frac{f(x+h)-f(x)}{h} \ &=\lim_{ h \to 0 } \frac{f(h)+hx}{h} \ &=\lim_{ h \to 0 } \frac{f(h)}{h}+x \ &=x+1 \end{align}

\begin{align} f(x)= \frac{1}{2}x^{2}+x\quad(\because f(0)=0) \ \therefore f(1)= \frac{3}{2} \end{align}

### 2.9 1. 좌극한은 0으로 수렴, 우극한은 -1~1에서 진동하므로 극한값이 존재하지 않는다. 2. $x\geq 0$인 경우 1) $x\geq \frac{1}{\pi}$인 경우 $0<\frac{1}{x}\leq\pi$이므로 $0\leq f_{1,1}(x)$ 2) $f_{1,1}\in\left( \left[ 0, \frac{1}{\pi} \right] \right)$인 경우 최대 최소 정리에 의해 $f_{1,1}(x_{1})\leq f_{1,1}(x)\leq f_{1,1}(x_{2}), x\in\left[ 0,\frac{1}{\pi} \right]$를 만족하는 $x_{1},x_{2}$가 존재한다. ($x\geq \frac{1}{\pi}$인 경우 항상 양수인 반면, $\left[ 0, \frac{1}{\pi} \right]$에서는 음수가 될 수 있기 때문에 이 구간 내에 최솟값이 존재한다) 3. 극한값 $\lim_{ n \to \infty }$