Identifying Discontinuities · Physics Playground

Example 1 from Precalculus, 12.3 Continuity

Identifying Discontinuities

Identify all discontinuities for the following functions as either a jump or a removable discontinuity.

ⓐ $f (x) = \frac{x^{2} - 2 x - 15}{x - 5}$
ⓑ $g (x) = {\begin{array}{l} x + 1, & x < 2 \\ - x, & x \geq 2 \end{array}$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. ⓐ
  Notice that the function is defined everywhere except at $x = 5.$
  
  Thus, $f (5)$ does not exist, Condition 2 is not satisfied. Since Condition 1 is satisfied, the limit as $x$ approaches 5 is 8, and Condition 2 is not satisfied.This means there is a removable discontinuity at $x = 5.$
2. ⓑ
  Condition 2 is satisfied because $g (2) = - 2.$
  
  Notice that the function is a piecewise function, and for each piece, the function is defined everywhere on its domain. Let’s examine Condition 1 by determining the left- and right-hand limits as $x$ approaches 2.
  Left-hand limit: $\lim_{x \to 2^{-}} (x + 1) = 2 + 1 = 3.$ The left-hand limit exists.
  Right-hand limit: $\lim_{x \to 2^{+}} (- x) = - 2.$ The right-hand limit exists. But
  
  $\lim_{x \to 2^{-}} f (x) \neq \lim_{x \to 2^{+}} f (x) .$
  
  So, $\lim_{x \to 2} f (x)$ does not exist, and Condition 2 fails: There is no removable discontinuity. However, since both left- and right-hand limits exist but are not equal, the conditions are satisfied for a jump discontinuity at $x = 2.$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Identify all discontinuities for the following functions as either a jump or a removable discontinuity.

ⓐ $f (x) = \frac{x^{2} - 6 x}{x - 6}$
ⓑ $g (x) = {\begin{array}{l} \sqrt{x}, & 0 \leq x < 4 \\ 2 x, & x \geq 4 \end{array}$

How did it go?