Example 1
x→2lim(3x+1) 3×(2)+1=7 from sympy import symbols, limit
x = symbols('x')
expr = 3*x + 1
result = limit(expr, x, 2)
print(result)
Example 2
x→0limxsin(x) Solution: Using Taylor Series
Use the Taylor expansion of sin(x) around 0:
sin(x)=x−3!x3+5!x5−... So:
xsin(x)=xx−3!x3+5!x5−...=1−6x2+120x4−... Now take the limit:
x→0lim(1−6x2+120x4−...)=1 from sympy import sin
expr = sin(x)/x
result = limit(expr, x, 0)
print(result)
Example 3
x→1limx−1x2−1 Solution: Factor numerator:
x−1x2−1=x−1(x−1)(x+1)=x+1⇒x→1lim(x+1)=2 expr = (x**2 - 1)/(x - 1)
result = limit(expr, x, 1)
print(result)
Example 4
x→0limx21−cos(x) Solution: This is a known standard limit:
Expand cos(𝑥) using Taylor series:
cos(x)=1−2!x2+4!x4−... So:
1−cos(x)=1−(1−2!x2+4!x4−...)=2!x2−4!x4+... Substitute into the limit:
x→0lim(x21−cos(x))=x→0lim(x22!x2−4!x4+...) Simplify the expression:
x→0lim(2!1−4!x2+...)=21 from sympy import cos
expr = (1 - cos(x))/x**2
result = limit(expr, x, 0)
print(result)
Example 5
x→∞lim2x2+75x2+5 x→∞lim2x2+75x2+5=x→∞lim2+x275+x25=25 from sympy import oo
expr = (5*x**2 + 3)/(2*x**2 + 7)
result = limit(expr, x, oo)
print(result)
Example 6
x→0lim∣x∣x Solution:
Left ≠ Right → Limit does not exist
from sympy import Abs
expr = x / Abs(x)
left_limit = limit(expr, x, 0, dir='-')
right_limit = limit(expr, x, 0, dir='+')
print(f"Left: {left_limit}, Right: {right_limit}")
Keywords
limit
, calculus
, trigonometric limits
, Taylor series
, squeeze theorem
, sin(x)/x
, fundamental limit
, L'Hôpital's rule
, limit definition
, mathematical proof
, nerd cafe
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