Removable Discontinuity

Core Definition

A function $f(x)$ has a removable discontinuity at $x = c$ if the two-sided limit $\lim_{x \to c} f(x) = L$ exists as a finite real number, but either:

$f(c)$ is undefined (a hole in the graph), OR
$f(c) \neq L$ (the point is plotted at a different height).

Concept Explanation

A removable discontinuity is colloquially referred to as a hole in the graph.

        y ^
          |          o  (c, L)  <-- approached limit L
        L + - - - - o
          |        /
        f(c)+    * (c, f(c))   <-- isolated function value
          |    /
          +---+----------> x
              c

It is called removable because the discontinuity can be completely eliminated (making the function continuous) simply by defining or redefining the single point $f(c)$ to match the limit value:

f_{\text{continuous}}(c) = \lim_{x \to c} f(x) = L

Algebraic Hallmarks

In algebraic formulas, removable discontinuities almost always arise from the indeterminate form $\frac{0}{0}$ :

\frac{P(x)}{Q(x)} = \frac{(x - c) \cdot A(x)}{(x - c) \cdot B(x)}

The factor

(x - c)

causes division by zero at

x = c

, making

f(c)

undefined. However, because a limit evaluates inputs near

c

(

x \neq c

), the common zero factor

(x - c)

can be canceled, revealing the true limit

L = \frac{A(c)}{B(c)}

Example Problems

Example Problem 1: Factoring to Find the Hole

Find the coordinates of the removable discontinuity (hole) for the function:

f(x) = \frac{x^2 - 9}{x - 3}

Solution

Step 1: Check direct substitution

f(3) = \frac{3^2 - 9}{3 - 3} = \frac{0}{0} \implies \text{undefined at } x = 3

Step 2: Evaluate the limit using factoring

\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6

Conclusion: The two-sided limit exists and equals $6$ . The graph of $f(x)$ is the straight line $y = x + 3$ with a removable discontinuity (hole) at $(3, 6)$ .

Example Problem 2: Removable Discontinuity with an Isolated Point

Analyze the discontinuity of the function:

g(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & : x \neq 1 \\ 5 & : x = 1 \end{cases}

Solution

Step 1: Evaluate the limit as $x \to 1$

\lim_{x \to 1} g(x) = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 2

Step 2: Compare limit with function value

Limit: $\lim_{x \to 1} g(x) = 2$
Function value: $g(1) = 5$

Step 3: Classify
Because $\lim_{x \to 1} g(x) \neq g(1)$ ( $2 \neq 5$ ), the function is discontinuous at $x = 1$ . Because the limit exists ( $L = 2$ ), this is a removable discontinuity. To make $g(x)$ continuous, redefine $g(1) = 2$ .

Connections & References

Parent Concepts: Discontinuities, Continuity
Related Concepts: Factoring Second Order Polynomials, Multiplying by Conjugates, Quotient Law
Course Links: Calc-1 Session 02 | Calc-1 Session 03