Limits

Limits#

Revised

14 Jun 2023

Programming Environment#

Hide code cell source 
import numpy             as np
import pandas            as pd
import matplotlib        as mpl
import matplotlib.pyplot as plt
from   matplotlib           import gridspec
from   mpl_toolkits.mplot3d import axes3d
from   ipywidgets           import interactive
plt.style.use('ggplot');

import sympy as smp
from   sympy import *

import plotly
import plotly.figure_factory as ff
import plotly.graph_objects  as go

from IPython.display import display, Math

from   datetime import datetime as d
import locale                   as l
import platform                 as p
import sys                      as s

pad = 20
print(f"{'Executed'.upper():<{pad}}: {d.now()}")
print()
print(f"{'Platform'        :<{pad}}: "
      f"{p.mac_ver()[0]} | "
      f"{p.system()} | "
      f"{p.release()} | "
      f"{p.machine()}")
print(f"{''                :<{pad}}: {l.getpreferredencoding()}")
print()
print(f"{'Python'          :<{pad}}: {s.version}")
print(f"{''                :<{pad}}: {s.version_info}")
print(f"{''                :<{pad}}: {p.python_implementation()}")
print()
print(f"{'Matplotlib'      :<{pad}}: {mpl   .__version__}")
print(f"{'NumPy'           :<{pad}}: {np    .__version__}")
print(f"{'Pandas'          :<{pad}}: {pd    .__version__}")
print(f"{'Plotly'          :<{pad}}: {plotly.__version__}")
print(f"{'SymPy'           :<{pad}}: {smp   .__version__}")

EXECUTED            : 2024-05-21 15:45:35.678137

Platform            : 14.4.1 | Darwin | 23.4.0 | arm64
                    : UTF-8

Python              : 3.11.9 | packaged by conda-forge | (main, Apr 19 2024, 18:34:54) [Clang 16.0.6 ]
                    : sys.version_info(major=3, minor=11, micro=9, releaselevel='final', serial=0)
                    : CPython

Matplotlib          : 3.8.4
NumPy               : 1.26.4
Pandas              : 2.2.2
Plotly              : 5.21.0
SymPy               : 1.12

\( \begin{aligned} 1=\lim_{\Delta x\to0}\frac{\sin\Delta x}{\Delta x} \end{aligned} \)

\( \begin{aligned} &\frac{2}{\sin\theta}\left(\frac{\tan\theta}{2}\ge\frac{\theta}{2}\ge\frac{\sin\theta}{2}\right) \iff \frac{1}{\cos\theta}\ge\frac{\theta}{\sin\theta}\ge1 \iff \cos\theta\le\frac{\sin\theta}{\theta}\le1 \\ &\lim_{\theta\to0}\cos\theta\le\lim_{\theta\to0}\frac{\sin\theta}{\theta}\le\lim_{\theta\to0}1 \implies 1\le\lim_{\theta\to0}\frac{\sin\theta}{\theta}\le1 \end{aligned} \)

\(\begin{aligned}\lim_{x\to\pi} \sin\left(\frac{x}{2} + \sin(x)\right)\end{aligned}\)#

\( \begin{aligned} \lim_{x\to\pi} \sin\left(\frac{x}{2} + \sin(x)\right) \end{aligned} \)

x = smp.symbols('x')

smp.limit(
  e   = smp.sin(x/2 + smp.sin(x)), # expression, the limit of which is to be taken
  z   = x,                         # symbol representing the variable in the limit
  z0  = smp.pi,                    # the value toward which `z` tends
  dir = '+-',                      # the limit is bidirectional
)
\[\displaystyle 1\]

\(\begin{aligned}\lim_{x \to 0^+} \frac{2e^\frac{1}{x}}{e^\frac{1}{x} + 1}\end{aligned}\)#

\( \begin{aligned} \lim_{x \to 0^+} \frac{2e^\frac{1}{x}}{e^\frac{1}{x} + 1} \end{aligned} \)

x = smp.symbols('x')

smp.limit(2 * smp.exp(1/x) / (smp.exp(1/x) + 1), x, 0, dir='+')
\[\displaystyle 2\]

\(\begin{aligned}\lim_{x \to 0^-} \frac{2e^\frac{1}{x}}{e^\frac{1}{x} + 1}\end{aligned}\)#

\( \begin{aligned} \lim_{x \to 0^-} \frac{2e^\frac{1}{x}}{e^\frac{1}{x} + 1} \end{aligned} \)

x = smp.symbols('x')

smp.limit(2 * smp.exp(1/x) / (smp.exp(1/x) + 1), x, 0, dir='-')
\[\displaystyle 0\]

\(\begin{aligned}\lim_{x \to \infty} \frac{\cos(x) - 1}{x}\end{aligned}\)#

\( \begin{aligned} \lim_{x \to \infty} \frac{\cos(x) - 1}{x} \end{aligned} \)

x = smp.symbols('x')

smp.limit((smp.cos(x) - 1)/x, x, smp.oo, dir='+-')
\[\displaystyle 0\]

Terms#

  • [W] Limit

Bibliography#

  • [Y] Mr. P Solver. (26 May 2021). “1st Year Calculus, But in PYTHON”. YouTube.

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