sympy – continuity test of functions

All functions I used here from Chapter 1 Exercises on Calculus and Analytic Geometry (9th Edition) by George B. Thomas, Ross L. Finney.

y = \frac{1}{x - 2} - 3 x
It has discontinuity at 2 of x. Smooth graph of a function means it has continuity in the given interval. Any gap or hole on a graph is discontinuous.

y1 = 1/(x - 2) - 3*x
plot(y1, (x, -2, 6), ylim=(-20, 20))

e1.5-13

y = \frac{x + 1}{x^{2} - 4 x + 3}

y2 = (x + 1) / (x**2 - 4*x + 3)
plot(y2, (x, 0, 4), ylim=(-10, 10))

e1.5-15

y = \left\lvert{x - 1}\right\rvert + \sin{\left (x \right )}

y17 = abs(x - 1) + sin(x)
plot(y17, (x, -5, 5), ylim=(-5, 5))

e1.5-17

y = \frac{1}{x} \cos{\left (x \right )}

y19 = (cos(x))/x 
plot(y19, (x, -10, 10), ylim=(-5, 5))

e1.5-19

y = \csc{\left (2 x \right )}

y21 = csc(2*x)
plot(y21, (x, 0, 6), ylim=(-3, 3))

e1.5-21

Finding limit of functions.

Can SymPy do it like Maxima does?

Find the limit of \sin{\left (x - \sin{\left (x \right )} \right )} function when x approaches to Pi.

y29 = sin(x - sin(x))
limit(y29, x, pi)

The output is 0.

Find the limit of y \sec^{2}{\left (y \right )} - \tan^{2}{\left (y \right )} - 1 function when y approaches to 1.

y = symbols('y')
y31 = y*(sec(y))**2 - (tan(y)**2) - 1
expr1 = trigsimp(y31)

It returns \frac{y - 1}{\cos^{2}{\left (y \right )}}

limit(expr1, y, 1)
Out[54]: 0
sec(0) 
Out[55]: 1

1 is the right answer.

Find the limit of \cos{\left (\frac{\pi}{\sqrt{- 3 \sec{\left (2 x \right )} + 19}} \right )} function when x approaches to 0.

y33 = cos(pi / (sqrt(19 - 3*sec(2*x))))
limit(y33, x, 0)

The output is \frac{\sqrt{2}}{2}

About janpenguin

Email: janpenguin [at] riseup [dot] net Every content on the blog is made by Free and Open Source Software in GNU/Linux.
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