미적분 이야기_CH1

1

1.  x=1에서 f(x)=x²의 접선의 기울기를 추정하기 위해
(a)(1,1), (2,4)를 통과하는 외선의 기울기?
(b)(1,1), (3/2, 9/4)를 통과하는 외선의 기울기?
(c)x=1에서의 접선의 기울기는 어떤 외선의 기울기에 근접할까요.

x=symbols('x')
f=x**2
f
x²
#(a)
a_slope=(1-4)/(1-2)
a_slope
3.0
#(b)
b_slope=(1-9/4)/(1-3/2)
b_slope
2.5
#(c)
limit(f, x, 1)
1

2

시간(t)에 따른 거리에 대한 식 s(t)=-16t²+64로부터 t=0.5초에서의 순간속도?

x, h=symbols('x, h')
f=-16*x**2+64
f2=(f.subs(x, x+h)-f)/h
f2
$\frac{16 x^{2} - 16 \left(h + x\right)^{2}}{h}$

v_avg=limit(f2, h, 0)
v_avg
−32x

v_avg.subs(x, 0.5)
−16.0

6

다음 극한의 연속 또는 불연속을 결정하세요?
(1)$\lim\limits_{x \to 2} \large \frac{x-3}{x(x-2)}$

x, h=symbols('x, h')
f=(x-3)/(x*(x-1))
f
$\frac{x - 3}{x \left(x - 1\right)}$
limit(f, x, 1, "-")
∞
limit(f, x, 1, "+")
−∞

(2)$\lim\limits_{x \to 1} \large \frac{x+3}{(x-1)^2}$

x, h=symbols('x, h')
f=(x+2)/(x-1)**2
f
$\frac{x + 2}{\left(x - 1\right)^{2}}$
limit(f, x, 1, "-")
∞
limit(f, x, 1, "+")
∞

(3)$\frac{x-1}{x^2+2x}=\frac{x-1}{x(x+2)}$
x=-2, 0에서 분모가 0이 됩니다. 그러므로 $(-\infty, -2)\; \cup \;(2, 0)\; \cup \;(0, \infty)$구간에서 연속

7

다음 극한을 계산하세요?
극한 계산에서 함수들은 가장 간단한 형태로 전환후에 실시합니다.

(1)$\lim\limits_{x \to 3} \large \frac{x^2-3x}{x^2-5x-3}$

x=symbols('x')
f=(x**2-3*x)/(2*x**2-5*x-3)
f
$\frac{x^{2} - 3 x}{2 x^{2} - 5 x - 3}$
f1=simplify(f)
f1
$\frac{x}{2 x + 1}$
limit(f1, x, 3)
$\frac{3}{7}$
limit(f, x, 3)
$\frac{3}{7}$

(2)$\lim\limits_{ x \to -1} \large \frac{\sqrt{x + 2} - 1}{x + 1}$

$ \begin{matrix}\frac{\sqrt{x + 2} - 1}{x + 1}&=\frac{(\sqrt{x + 2} -1)(\sqrt{x + 2} +1)}{(x + 1)(\sqrt{x + 2} +1)}\\&=\frac{x+2-1}{(x + 1)(\sqrt{x + 2} +1)}\\&=\frac{1}{\sqrt{x + 2} +1}\end{matrix}$

$\lim\limits_{x \to -1} \large \frac{1}{\sqrt{x + 2} +1}=\frac{1}{2}$

x=symbols('x')
f=(sqrt(x+2)-1)/(x+1)
limit(f, x, -1)
$\frac{1}{2}$

(3)$\lim\limits_{ x \to 5} \large \frac{\sqrt{x - 1} - 2}{x - 5}$

$ \begin{matrix}\frac{\sqrt{x - 1} - 2}{x - 5}& =\frac{(\sqrt{x - 1} -2)(\sqrt{x - 1} +2)}{(x-5)(\sqrt{x- 1}+2)}\\ &=\frac{x-1-4}{(x -5)(\sqrt{x-1} +2)}\\ &=\frac{1}{\sqrt{x-1} +2}\end{matrix}$

$\lim\limits_{x \to 5} \large \frac{1}{\sqrt{x-1} +2}=\frac{1}{4}$

x=symbols('x')
f=(sqrt(x-1)-2)/(x-5) limit(f, x, 5)
$\frac{1}{4}$

(4)$\lim\limits_{ x \to 1} \large \frac{\frac{1}{x + 1}-\frac{1}{2}}{x - 1}$

$ \begin{matrix}\frac{\frac{1}{x + 1}-\frac{1}{2}}{x - 1}&=\frac{\frac{-(x-1)}{2(x+1)}}{x-1}\\&=\frac{-(x-1)}{2(x+1)(x-1)}\\&=\frac{-1}{2(x+1)}\end{matrix}$

$\lim\limits_{x \to 1} \large -\frac{1}{2(x+1)}=-\frac{1}{4}$

x=symbols('x')
f=(1/(x+1) -1/2)/(x-1)
limit(f, x, 1)
$−\frac{1}{4}$

8

다음 함수의 점근선을 결정하세요?

(1)$\frac{3x-1}{2x+5}$

$\begin{align}\lim\limits_{x \to \infty}\frac{3-\frac{1}{x}}{2+\frac{5}{x}}=\frac{3}{2}\\ \lim\limits_{x \to -\infty}\frac{3-\frac{1}{x}}{2+\frac{5}{x}}=\frac{3}{2}\end{align}$

(2)$\frac{3x^2-2x}{4x^3-5x+7}$

$\begin{align}\lim\limits_{x \to \infty}\frac{\frac{3}{x}-\frac{2}{x^2}}{4-\frac{5}{x^2}+\frac{7}{x^3}}=0\\\lim\limits_{x \to -\infty}\frac{\frac{3}{x}-\frac{2}{x^2}}{4-\frac{5}{x^2}+\frac{7}{x^3}}=0\end{align}$

9

다음을 계산하세요?

$(1) \lim\limits_{x \to 0} x^2 cos\left(\frac{1}{x}\right)$
$t=\frac{1}{x},\quad x \rightarrow 0 \Rightarrow t \rightarrow \infty$
$\begin{align}\lim\limits_{x \to 0} x^2 cos\left(\frac{1}{x}\right)&=\lim\limits_{t \to \infty}\frac{1}{t^2}cos(t)\\&=0\end{align}$

x=symbols('x')
f=x**2*cos(1/x)
limit(f, x, 0)
0

$(2)\quad\begin{align}\lim\limits_{x \to 0} \frac{tan(x)}{x}&=\lim\limits_{x \to 0}\frac{1}{x} \frac{sin(x)}{cos(x)}\\&=\lim\limits_{x \to 0}\frac{sin(x)}{x}\lim\limits_{x \to 0}\frac{1}{cos(x)}\\&=1\end{align}$

x=symbols('x')
f=tan(x)/x
limit(f, x, 0)
1

$(3)\quad\lim\limits_{x \to \infty} xsin\left(\frac{1}{x}\right)$
$\frac{1}{x}=t, \quad x \to \infty \Rightarrow t \to 0$
$\begin{align}\lim\limits_{x \to \infty} xsin\left(\frac{1}{x}\right)&=\lim\limits_{t \to 0}\frac{sin(t)}{t}\\&=1\end{align}$

x=symbols('x')
f=x*sin(1/x)
limit(f, x, oo)
1

$(4)\quad\begin{align}\lim\limits_{x \to 0}\frac{x+2sin(x)}{sin(x)cos(x)}&=\lim\limits_{x \to 0}\frac{x}{sin(x)cos(x)}+\lim\limits_{x \to 0}\frac{2sin(x)}{sin(x)cos(x)}\\ &=\lim\limits_{x \to 0}\frac{1}{\frac{sin(x)}{x}}\lim\limits_{x \to 0}\frac{1}{cos(x)}+\lim\limits_{x \to 0}\frac{2}{cos(x)}\\ &=1 \cdot 1+ 2\\&=3\end{align}$

x=symbols('x')
f=(x+2*sin(x))/(sin(x)*cos(x))
limit(f, x, 0)
3

9

다음을 계산하세요?

$(1) \lim\limits_{x \to 0} x^2 cos\left(\frac{1}{x}\right)$
$t=\frac{1}{x},\quad x \rightarrow 0 \Rightarrow t \rightarrow \infty$
$\begin{align}\lim\limits_{x \to 0} x^2 cos\left(\frac{1}{x}\right)&=\lim\limits_{t \to \infty}\frac{1}{t^2}cos(t)\\&=0\end{align}$

x=symbols('x')
f=x**2*cos(1/x)
limit(f, x, 0)
0

$(2)\quad\begin{align}\lim\limits_{x \to 0} \frac{tan(x)}{x}&=\lim\limits_{x \to 0}\frac{1}{x} \frac{sin(x)}{cos(x)}\\&=\lim\limits_{x \to 0}\frac{sin(x)}{x}\lim\limits_{x \to 0}\frac{1}{cos(x)}\\&=1\end{align}$

x=symbols('x')
f=tan(x)/x
limit(f, x, 0)
1

$(3)\quad\lim\limits_{x \to \infty} xsin\left(\frac{1}{x}\right)$
$\frac{1}{x}=t, \quad x \to \infty \Rightarrow t \to 0$
$\begin{align}\lim\limits_{x \to \infty} xsin\left(\frac{1}{x}\right)&=\lim\limits_{t \to 0}\frac{sin(t)}{t}\\&=1\end{align}$

x=symbols('x')
f=x*sin(1/x)
limit(f, x, oo)
1

$(4)\quad\begin{align}\lim\limits_{x \to 0}\frac{x+2sin(x)}{sin(x)cos(x)}&=\lim\limits_{x \to 0}\frac{x}{sin(x)cos(x)}+\lim\limits_{x \to 0}\frac{2sin(x)}{sin(x)cos(x)}\\ &=\lim\limits_{x \to 0}\frac{1}{\frac{sin(x)}{x}}\lim\limits_{x \to 0}\frac{1}{cos(x)}+\lim\limits_{x \to 0}\frac{2}{cos(x)}\\ &=1 \cdot 1+ 2\\&=3\end{align}$

x=symbols('x')
f=(x+2*sin(x))/(sin(x)*cos(x))
limit(f, x, 0)
3

10

다음을 계산하세요?

$(1)\lim\limits_{n \to \infty} \large n \cdot ln\left(1+\frac{x}{n}\right)$

$u=\frac{x}{n}, \quad n \to \infty \Rightarrow u \to 0$
$\begin{align}\lim\limits_{n \to \infty} \large n \cdot \left(1+\frac{x}{n}\right)&=\lim\limits_{u \to 0} \frac{x}{u} \cdot ln(1+u)\\&=x\lim\limits_{u \to 0} ln(1+u)^{\frac{1}{u}}\\&=x \end{align}$

n, x=symbols('n x')
f=n*ln(1+x/n)
limit(f, n, oo)
x

$(2)\begin{align}\lim\limits_{n \to \infty} \frac{2x-7}{\sqrt{9x^2+7x+2}}&=\lim\limits_{n \to \infty}\frac{2-\frac{7}{x}}{\sqrt{9+\frac{7}{x}+\frac{2}{x^2}}}\\&=\frac{2}{3} \end{align}$

x=symbols('x')
f=(2*x-7)/(sqrt(9*x**2+7*x+2))
limit(f, x, oo)
$\frac{2}{3}$

(3) $\begin{align}\lim\limits_{x \to \infty} \sqrt{x^{2} + 5 x + 9}- \sqrt{x^{2} - 4 x + 2} &=\lim\limits_{x \to \infty}\frac{(x^{2} + 5 x + 9)-(x^{2} - 4 x + 2)}{\sqrt{x^{2} + 5 x + 9}+\sqrt{x^{2} - 4 x + 2}}\\&=\lim\limits_{x \to \infty}\frac{ 9x + 7}{\sqrt{x^{2} + 5 x + 9}+\sqrt{x^{2} - 4 x + 2}}\\&=\lim\limits_{x \to \infty}\frac{ 9 + \frac{7}{x}}{\sqrt{1 + \frac{5}{x} +\frac{9}{x^2}}+\sqrt{1 - \frac{4}{x}+\frac{2}{x^2}}}\\&=\frac{9}{2}\end{align}$

x=symbols('x')
f=sqrt(x**2+5*x+9)-sqrt(x**2-4*x+2)
limit(f, x, oo)
$\frac{9}{2}$