1. Definition of Continuity
In simple terms, a function is **continuous** if you can draw its graph without lifting your pencil from the paper. There are no holes, jumps, or gaps.
More formally, a function \( f \) is continuous at a point \( c \) if and only if all three of the following conditions are met:
- \( f(c) \) is defined.
- \( \lim_{x \to c} f(x) \) exists.
- \( \lim_{x \to c} f(x) = f(c) \).
If any one of these three conditions fails, the function is said to be **discontinuous** at \( c \).
2. Types of Discontinuities
Discontinuities can be classified into two main types:
Removable Discontinuities
A **removable discontinuity** is a single point (a "hole" in the graph) where the limit exists, but the function's value is either undefined or doesn't match the limit. This can be "fixed" by redefining the function at that single point.
Example: \( f(x) = \frac{x^2 - 4}{x - 2} \) at \( x=2 \)
This function has a hole at \( x=2 \). The limit as \( x \to 2 \) is 4, but \( f(2) \) is undefined. The discontinuity is removable.
Non-Removable Discontinuities
A **non-removable discontinuity** cannot be fixed by redefining a single point. The limit at that point simply does not exist.
- **Jump Discontinuity:** The left-hand and right-hand limits exist but are not equal to each other. This is common in piecewise functions.
- **Infinite Discontinuity:** The function's value approaches positive or negative infinity as \( x \) approaches the point. This occurs at a vertical asymptote.
3. Check Yourself
Determine where each function is discontinuous and classify the type of discontinuity.
Question 1: \( f(x) = \frac{x^2 - x - 6}{x-3} \)
The function is undefined at \( x=3 \). By factoring the numerator, we get \( f(x) = \frac{(x-3)(x+2)}{x-3} = x+2 \) for \( x \neq 3 \). The limit as \( x \to 3 \) is \( 3+2 = 5 \). Since the limit exists but the function is undefined, there is a **removable discontinuity** at \( x=3 \).
Question 2: \( g(x) = \begin{cases} 2x+1 & \text{if } x < 1 \\ x^2 & \text{if } x \geq 1 \end{cases} \)
We check for continuity at \( x=1 \).
Left-hand limit: \( \lim_{x \to 1^-} (2x+1) = 2(1)+1 = 3 \)
Right-hand limit: \( \lim_{x \to 1^+} x^2 = (1)^2 = 1 \)
Since the left-hand limit and the right-hand limit are not equal, the overall limit does not exist. This is a **non-removable jump discontinuity** at \( x=1 \).
Question 3: \( h(x) = \frac{5}{x+4} \)
The function is undefined at \( x=-4 \). As \( x \) approaches -4, the denominator approaches 0, and the function's value approaches \( \infty \) or \( -\infty \). Since the limit is infinite, there is a **non-removable infinite discontinuity** (a vertical asymptote) at \( x=-4 \).