參考書籍：畢文斌, 毛悅悅. Python漫游數學王國[M]. 北京：清華大學出版社，2022.

1 無窮小與無窮大

無窮大在sympy中用兩個字母o表示無窮大，正無窮大為sy.oo，負無窮大為-sy.oo

import sympy as sy
x=sy.oo
print(1/x)

>>>0

${\lim }_{x\to 0^{?}}\frac{1}{x}$

x=sy.symbols('x')
print(sy.limit(1/x,x,0,dir='-'))

>>>-oo

2 極限運算法則

${\lim }_{x\to 3}\frac{x?3}{x^2?9}$

import sympy as sy
x=sy.symbols('x')
print(sy.limit((x-3)/(x**2-9),x,3,dir='+-'))

${\lim }_{x\to 1}\frac{2x?3}{x^2?5x+4}$

x=sy.symbols('x')
print(sy.limit((2*x-3)/(x**2-5*x+4),x,1,dir='-'))
print(sy.limit((2*x-3)/(x**2-5*x+4),x,1))

>>>-oo, oo 故趨于無窮時極限為無窮oo

${\lim }_{x\to \infty }\frac{3x^3+4X^2+2}{7x^3+5x^2?3}$

x=sy.symbols('x')
print(sy.limit((3*x**3+4*x**2+2)/(7*x**3+5*x**2-3),x,sy.oo,dir='-'))
print(sy.limit((3*x**3+4*x**2+2)/(7*x**3+5*x**2-3),x,-sy.oo,dir='+'))

>>>3/7,3/7 故趨于無窮時極限為3/7

當分子分母極限都不存在時，${\lim }_{x\to \infty }\frac{\sin x}{x}$

x=sy.symbols('x')
y=sy.sin(x)/x
print(sy.limit(y,x,sy.oo,dir='+'))
print(sy.limit(y,x,-sy.oo,dir='+'))

>>>0 , 0 故趨于無窮時極限為0

3 極限存在原則

eg1: ${\lim }_{x\to 0}\frac{\sin x}{x}$

import sympy as sy
x=sy.symbols('x')
lim=sy.limit(sy.sin(x)/x,x,0,dir='+-')
print(lim)

>>>1

eg2: ${\lim }_{x\to 0}\frac{\arcsin x}{\tan x}$

x=sy.symbols('x')
print(sy.limit(sy.asin(x)/sy.tan(x),x,0,dir='+-'))  #sy.asin()指arcsin函數

>>>1

eg3: ${\lim }_{x\to 0}\frac{1?\cos x}{x^2}$

x=sy.symbols('x')
print(sy.limit((1-sy.cos(x))/(x**2),x,0,dir='+-'))

>>>1/2

eg4: ${\lim }_{x\to 0}(1+x{)}^{\frac{1}{x}}$

x=sy.symbols('x')
lim=sy.limit((1+x)**(1/x),x,0,dir='+-')
print(lim)

>>>E

eg5: ${\lim }_{x\to \infty }(1+\frac{1}{x}{)}^x$

x=sy.symbols('x')
lim=sy.limit((1+1/x)**x,x,sy.oo,dir='-')
print(lim)
print(lim.round(3))
print(sy.limit((1+1/x)**x,x,-sy.oo))

>>>E, 2.718, E

eg6: 說明數列$\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}}$,···的極限存在

#用函數的遞歸機制定義數列
def a_complex_series(n):#退出條件if n<=0:return 2**0.5#一個函數如果調用自身，則這個函數就是一個遞歸函數return (2.0+a_complex_series(n-1))**0.5
#繪制前20個數的散點圖
import matplotlib.pyplot as plt
import numpy as np
x=[]
y=[]
for i in range(20):x.append(i)y.append(a_complex_series(i))
print(np.array(y))
plt.scatter(x,y)
plt.show()

>>>[1.41421356 1.84775907 1.96157056 1.99036945 1.99759091 1.99939764
1.9998494  1.99996235 1.99999059 1.99999765 1.99999941 1.999999851.99999996 1.99999999 2.         2.         2.         2.
2.         2.        ]

故極限為2

4 趨于無窮小的比較

eg1: ${\lim }_{x\to 0}\frac{\tan 2x}{\sin 5x}$

from sympy import limit,sin,cos,tan,symbols #從sympy中僅導入這幾個函數
x=symbols('x')
example_1=tan(2*x)/sin(5*x)
result=limit(example_1,x,0,dir='+-')
print(result)

>>>2/5

eg2: ${\lim }_{x\to 0}\frac{\sin x}{x^3+3x}$

x=symbols('x')
example_2=sin(x)/(x**3+3*x)
result=limit(example_2,x,0,dir='+-')
print(result)

>>>1/3

eg3: ${\lim }_{x\to 0}\frac{(1+x^2{)}^{1/3}?1}{\cos x?1}$

x=symbols('x')
example_3=((1+x**2)**(1/3)-1)/(cos(x)-1)
result=limit(example_3,x,0,dir='+-')
print(result)

>>>-2/3