Continuity and Discontinuity: Definitions and Classifications

Definition of Continuity at a Point

A function is continuous at a point $x = a$ if and only if the limit as $x$ approaches $a$ of the function equals the function's value at $a$ . Mathematically, this means: $\lim_{x \to a} f(x) = f(a)$ .
This implies three conditions must be met:
1. The limit $\lim_{x \to a} f(x)$ must exist.
2. The function value $f(a)$ must be defined.
3. These two values must be equal.

Types of Discontinuities

Jump Discontinuity

Condition: The limit as $x$ approaches $a$ from the left exists, and the limit as $x$ approaches $a$ from the right exists, but these two one-sided limits are not equal.
- $\lim{x \to a^-} f(x) = L1$ and $\lim{x \to a^+} f(x) = L2$ , where $L1 \neq L2$ .
Result: The overall limit $\lim_{x \to a} f(x)$ does not exist.
Characteristic: The graph of the function "jumps" from one value to another at $x = a$ .

Infinite Discontinuity

Condition: At least one of the one-sided limits is either positive or negative infinity.
- For example, $\lim{x \to a^-} f(x) = \pm \infty$ or $\lim{x \to a^+} f(x) = \pm \infty$ .
Note: If both one-sided limits go to $\pm \infty$ , it is still classified as an infinite discontinuity.
Characteristic: The graph approaches a vertical asymptote at $x = a$ .

Removable Discontinuity

Condition: The limit as $x$ approaches $a$ exists and is finite, but either:
1. The function value $f(a)$ is not defined.
2. The function value $f(a)$ is defined but not equal to the limit.
- $\lim_{x \to a} f(x) = L$ , where $L$ is a finite number, but $f(a)$ is undefined or $f(a) \neq L$ .
Characteristic: There is a "hole" in the graph at $x = a$ . It is called "removable" because the discontinuity can be fixed by redefining $f(a)=L$ .

More Severe Type of Discontinuity

Condition: At least one of the one-sided limits does not exist and is not infinite.
- For instance, if $\lim_{x \to a^+} f(x)$ does not exist (e.g., due to oscillation) and is not $\pm \infty$ , then it's a more severe type of discontinuity.
Characteristic: This indicates a more erratic behavior than the other types, where the function might oscillate infinitely or exhibit chaotic behavior near the point.

Left and Right Continuity

Left Continuous: A function is left continuous at $x = a$ if $\lim_{x \to a^-} f(x) = f(a)$ .
Right Continuous: A function is right continuous at $x = a$ if $\lim_{x \to a^+} f(x) = f(a)$ .
A function is continuous at a point if and only if it is both left and right continuous at that point.
It is not typically useful to discuss left or right continuity for removable discontinuities, as the focus is on whether the limit exists.

Solving Problems Involving Continuity

To classify continuity at a point, one must always check:
1. The left-hand limit ( $\lim_{x \to a^-} f(x)$ ).
2. The right-hand limit ( $\lim_{x \to a^+} f(x)$ ).
3. The function's value ( $f(a)$ ).

Example 1: Making a Piecewise Function Continuous

Problem: Given a piecewise function where each piece is a polynomial (continuous everywhere), find the value of a constant $'a'$ that makes the function continuous at a specific gluing point (e.g., $x = 2$ ).
Steps for Solution:
1. Identify the gluing point: This is where the function definition changes (e.g., $x = 2$ ).
2. Calculate the left-hand limit: For x < 2, using the function piece $f(x) = ax^2 + 2x$ .
 - $\lim_{x \to 2^-} (ax^2 + 2x) = a(2^2) + 2(2) = 4a + 4$ .
3. Calculate the right-hand limit: For x > 2, using the function piece $f(x) = x^3 - ax$ .
 - $\lim_{x \to 2^+} (x^3 - ax) = 2^3 - a(2) = 8 - 2a$ .
4. Set limits equal: For the limit to exist, the left-hand and right-hand limits must be equal.
 - $4a + 4 = 8 - 2a$
 - $6a = 4$
 - $a = \frac{4}{6} = \frac{2}{3}$ .
5. Verify function value: With $a = \frac{2}{3}$ , the function definition at $x = 2$ uses the second piece: $f(x) = x^3 - \frac{2}{3}x$
 - $f(2) = 2^3 - \frac{2}{3}(2) = 8 - \frac{4}{3} = \frac{24 - 4}{3} = \frac{20}{3}$ .
 - The limit (from step 4) would be $8 - 2(\frac{2}{3}) = 8 - \frac{4}{3} = \frac{20}{3}$ . Since the limit equals the value, the function is continuous at $x = 2$ when $a = \frac{2}{3}$ .
Conclusion: The function is continuous everywhere when $a = \frac{2}{3}$ .

Example 2: Classifying Discontinuity for a Piecewise Function

Problem: Classify the discontinuity of a given piecewise function at a specific point (e.g., $x = 0$ ), and determine if it's left or right continuous.
- $f(x) = \begin{cases} 1 + x^2 & \text{if } x < 0 \ 2 - x & \text{if } 0 < x < 2 \ x & \text{if } x \ge 2 \end{cases}$
Steps for Solution at $x = 0$ :
1. Calculate the left-hand limit: For x < 0, use $f(x) = 1 + x^2$ .
 - $\lim_{x \to 0^-} (1 + x^2) = 1 + 0^2 = 1$ .
2. Calculate the right-hand limit: For 0 < x < 2, use $f(x) = 2 - x$ .
 - $\lim_{x \to 0^+} (2 - x) = 2 - 0 = 2$ .
3. Compare one-sided limits: Since $1 \neq 2$ , the left-hand limit and right-hand limit are not equal. Therefore, $\lim_{x \to 0} f(x)$ does not exist.
4. Classify Discontinuity: Because the one-sided limits exist but are not equal, the function has a jump discontinuity at $x = 0$ .
5. Check function value: At $x = 0$ , the definition specifies the value based on the first piece or a separate definition, but here the teacher states $f(0)$ is defined from the $x \le 0$ part if the range for the first piece actually includes $0$ . Assuming the definition for x < 0 is used for $x \le 0$ by context, we infer $f(0) = 1 + 0^2 = 1$ . (In the transcript, the teacher states "Here's where $x$ is equal to zero. So you plug in zero and you get one." which seems implied by the first piece or an unstated part of the definition).
6. Determine left/right continuity:
 - Is it left continuous? $\lim_{x \to 0^-} f(x) = 1$ and $f(0) = 1$ . Since they are equal, the function is left continuous at $x = 0$ .
 - It cannot be right continuous because $\lim_{x \to 0^+} f(x) = 2$ which is not equal to $f(0) = 1$ .
Conclusion: The function has a jump discontinuity at $x = 0$ and is left continuous.

Example 3: Discontinuity Involving Absolute Value

Problem: Classify the discontinuity of $f(x) = \frac{1}{x-1} + \frac{1}{|x-1|}$ at $x = 1$ .
Steps for Solution:
1. Rewrite with piecewise definition of absolute value:
 - $|x-1| = \begin{cases} x-1 & \text{if } x-1 \ge 0 \implies x \ge 1 \ -(x-1) & \text{if } x-1 < 0 \implies x < 1 \end{cases}$ .
2. Simplify $f(x)$ :
 - For x > 1 (since the problem implicitly doesn't define for $x=1$ ), $f(x) = \frac{1}{x-1} + \frac{1}{x-1} = \frac{2}{x-1}$ .
 - For x < 1, $f(x) = \frac{1}{x-1} + \frac{1}{-(x-1)} = \frac{1}{x-1} - \frac{1}{x-1} = 0$ .
 - So, the simplified function is: $f(x) = \begin{cases} 0 & \text{if } x < 1 \ \frac{2}{x-1} & \text{if } x > 1 \end{cases}$ .
3. Calculate the left-hand limit: For x < 1, use $f(x) = 0$ .
 - $\lim_{x \to 1^-} 0 = 0$ .
4. Calculate the right-hand limit: For x > 1, use $f(x) = \frac{2}{x-1}$ .
 - Plug in $x = 1$ into the denominator: $2/(1-1) = 2/0$ , indicating an infinite limit.
 - Test a value slightly greater than $1$ (e.g., $1.1$ ): $\frac{2}{1.1-1} = \frac{2}{0.1} = 20$ . Since it's positive, $\lim_{x \to 1^+} \frac{2}{x-1} = +\infty$ .
5. Compare one-sided limits: The left-hand limit is $0$ , and the right-hand limit is $+\infty$ .
6. Classify Discontinuity: Since one of the one-sided limits is $+\infty$ , the function has an infinite discontinuity at $x = 1$ .
7. Check function value: $f(1)$ is undefined because the original expression would have $1-1=0$ in the denominator.
8. Determine left/right continuity: Since $f(1)$ does not exist, the function cannot be left or right continuous at $x = 1$ .
Conclusion: The function has an infinite discontinuity at $x = 1$ , and it is neither left nor right continuous.

Important Considerations

When checking continuity on a quiz, always provide the explicit calculations for left-hand limits, right-hand limits, and the function's value at the point in question.
The explanation of your findings is crucial for full credit, though showing the limit calculations is the bare minimum.
For removable discontinuities, the limit exists, and it's possible to redefine the function at that single point to make it continuous (e.g., setting $f(a) = \lim_{x \to a} f(x)$ ).
A function cannot be both removable and left/right continuous at the point of discontinuity simultaneously because if it were, it would imply continuity, which contradicts the definition of a discontinuity.