A removable discontinuity on a graph is a point where a function is not defined—or where its value does not match the limit—but the gap can be “filled in” by redefining the function at that single point so that the graph becomes continuous. Basically, the break is only apparent; the function behaves smoothly everywhere else, and the limit from both sides exists and is equal. Understanding this concept is essential for calculus students because it clarifies how limits work, how functions can be simplified, and why certain algebraic manipulations are permissible.
Understanding Discontinuities in Functions
Before diving into the specifics of a removable discontinuity, it helps to recall what a discontinuity is in general. A function f(x) is said to be continuous at a point x = a if three conditions hold:
- f(a) is defined.
- The limit (\displaystyle \lim_{x \to a} f(x)) exists.
- The limit equals the function value: (\displaystyle \lim_{x \to a} f(x) = f(a)).
If any of these conditions fails, the function is discontinuous at x = a. Discontinuities come in several flavors—removable, jump, and infinite—each with its own graphical signature and algebraic cause Nothing fancy..
Definition of a Removable Discontinuity
A removable discontinuity occurs at x = a when:
- The limit (\displaystyle \lim_{x \to a} f(x)) exists (and is finite).
- Either f(a) is undefined, or f(a) is defined but not equal to that limit.
Because the limit exists, the “hole” in the graph can be removed by assigning f(a) the value of the limit. After this redefinition, the function satisfies all three continuity conditions at x = a It's one of those things that adds up..
In symbols, if
[ L = \lim_{x \to a} f(x) \quad \text{exists and is finite}, ]
and either f(a) is undefined or f(a) ≠ L, then x = a is a removable discontinuity. The term “removable” reflects the fact that we can remove the discontinuity by redefining the function at that single point.
Visual Identification on a Graph
Graphically, a removable discontinuity appears as a hole (often drawn as an open circle) in an otherwise smooth curve. The function approaches the same y‑value from the left and the right, but the point itself is missing or mismatched Small thing, real impact..
- Open circle: Indicates the point that is not part of the original function.
- Approaching curve: Shows that the function values get arbitrarily close to a specific y‑value as x approaches the hole from either side.
- Possible filled point: Sometimes a separate, solid dot is plotted elsewhere to show the actual defined value of f(a) (if it exists but differs from the limit).
If you were to lift your pencil and trace the graph, you would need to lift it only at the hole; otherwise the pen could move continuously across the rest of the curve And that's really what it comes down to..
Algebraic Explanation: Limits and Factor Cancellation
Removable discontinuities frequently arise from rational functions where a factor in the numerator and denominator cancels out, leaving a domain restriction. Consider a function
[ f(x) = \frac{(x-2)(x+3)}{x-2}. ]
At first glance, the function seems undefined at x = 2 because the denominator becomes zero. Still, for all x ≠ 2 we can cancel the common factor (x-2), obtaining the simplified expression
[ f(x) = x + 3 \quad \text{for } x \neq 2. ]
The limit as x approaches 2 is straightforward:
[ \lim_{x \to 2} f(x) = \lim_{x \to 2} (x+3) = 5. ]
Since the limit exists and equals 5, but f(2) is undefined (the original formula gives division by zero), the point (2, 5) is a removable discontinuity. Graphically, the function looks exactly like the line y = x + 3 except for a hole at (2, 5) Most people skip this — try not to..
If we redefine the function to fill the hole—say, let
[ g(x) = \begin{cases} \frac{(x-2)(x+3)}{x-2}, & x \neq 2\ 5, & x = 2 \end{cases} ]
then g(x) is continuous everywhere, including at x = 2.
Examples
Example 1: Simple Rational Hole
[ h(x) = \frac{x^2 - 9}{x - 3}. ]
Factor the numerator: (x-3)(x+3). Cancel (x-3) for x ≠ 3, leaving h(x) = x + 3 (with a hole at x = 3).
[ \lim_{x \to 3} h(x) = 6, \quad h(3) \text{ undefined}. ]
Thus, (3, 6) is a removable discontinuity Small thing, real impact..
Example 2: Piecewise Function with Mismatched Value
[ p(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0\ 1, & x = 0 \end{cases} ]
We know (\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1). Consider this: since p(0) is already defined as 1, the function is actually continuous at zero—no discontinuity. If we instead defined p(0) = 2, then we would have a removable discontinuity at (0, 1) because the limit is 1 but the function value is 2.
Example 3: Radical Expression
[ r(x) = \frac{\sqrt{x+1} - 1}{x}. ]
Direct substitution at x = 0 yields 0/0. Rationalizing the numerator:
[ r(x) = \frac{(\sqrt{x+1} - 1)(\sqrt{x+1} + 1)}{x(\sqrt{x+1} + 1)} = \frac{x}{x(\sqrt{x+1} + 1)} = \frac{1}{\sqrt{x+1} + 1}, \quad x \neq 0. ]
Now (\displaystyle \lim_{x \to 0} r(x) = \frac{1}{2
