Continuity, question 12

Topic: Continuity
Difficulty: Standard
Suggested time: 4 min
Marks: 2
Theory: UBC Math 100 §2.3 CLP-1 §1.6Key results

Find and classify every discontinuity of the function below.

g(x) = \frac{x^{3} - x}{x^{3} - 2x^{2} + x}

Removable discontinuities at x = 0 and x = 1.
A removable discontinuity at x = 0 and an infinite discontinuity at x = 1.
Infinite discontinuities at x = 0 and x = 1.
An infinite discontinuity at x = 0 and a removable discontinuity at x = 1.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §2.3 CLP-1 §1.6Key results

Worked solution

Strategy

A rational function is continuous except where its denominator is zero. Factor, cancel what you can, and look at what is left at each excluded point: a finite limit means a hole, a surviving zero in the denominator means a vertical asymptote.

Working

$x^{3} - x = x(x - 1)(x + 1), \qquad x^{3} - 2x^{2} + x = x(x - 1)^{2}$
Factor fully. The denominator is zero only at x = 0 and x = 1, and it contains x − 1 twice.
$g(x) = \frac{x(x - 1)(x + 1)}{x(x - 1)^{2}} = \frac{x + 1}{x - 1} \quad (x \neq 0,\ 1)$
Cancel x and one copy of x − 1. g is still undefined at 0 and 1.
$\lim_{x \to 0} g(x) = \frac{0 + 1}{0 - 1} = -1$
A finite limit at a point where g is undefined: a hole, so the discontinuity at 0 is removable.
$\lim_{x \to 1^{-}} \frac{x + 1}{x - 1} = -\infty, \qquad \lim_{x \to 1^{+}} \frac{x + 1}{x - 1} = +\infty$
One copy of x − 1 survives in the denominator while the numerator tends to 2: an infinite discontinuity.

Answer

\text{removable at } x = 0, \quad \text{infinite at } x = 1

Check

g(0.01) ≈ −1.02 stays near −1, but g(0.99) = −199 and g(1.01) = 201: a hole at 0 and a vertical asymptote at 1.

Where this goes wrong

Calling x = 1 removable because a factor x − 1 cancels. The denominator holds (x − 1)², so one copy survives the cancelling and the function still blows up there.

Leans on

Factoring polynomials, from Functions, graphs and inverses

Try a similar question

Classify the discontinuities of the function below.

h(x) = \frac{x^{2} - x - 2}{x^{2} - 4}

My working

Saved in this browser only.

Continuity practice question 12