Algebraic Simplification
Author: Jamie Mulholland

Refer to example 7(a) in section 2.3 of the the class notes.

The original function has a hole at x=-1 (i.e. -1 is not in the domain), this is illustrated in the diagram below.  However, when we simplify the function and we obtain a function which is equivalent to f, for x not equal to-1, and is itself defined at -1.  In other words, we remove the hole.  This allows us to compute the limit of f at -1 by computing the corresponding value of g at -1.

Move the slider in the diagram below to remove the hole.

hole

no hole

\lim_{x \to -1} \frac{x+1}{x^3+1} = \lim_{x\to -1} \frac{1}{x^2-x+1} = 1

f(x) = (x + 1) / (x^(3) + 1)
g(x) = 1 / (x^(2) - x + 1)
Hole: (1, g(1))
T1: "f(x) = " + f
T2: "g(x) =" + g
toggle = 0 (slider from 0 to 1, step 1)