Exploring Types of Discontinuities - AP Calculus BC Study Guide

The Core Idea: Exploring Types of Discontinuities

A function is continuous at a point if the limit at that point exists, the function value at that point exists, and these two values are equal. A discontinuity occurs when one or more of these conditions fail. This topic focuses on classifying the different reasons why a function might be discontinuous at a specific point, $x=c$.

The classification of a discontinuity is determined entirely by the behavior of the function as $x$ approaches $c$ from the left and the right. By analyzing the one-sided limits, we can distinguish between discontinuities that are single, isolated points (removable), those that represent a sudden jump in function values (jump), and those where the function values increase or decrease without bound (due to a vertical asymptote). Understanding these classifications is fundamental to describing the behavior of functions.

Key Definitions

The type of a discontinuity at a point $x=c$ is defined by the existence and relationship of the one-sided limits and the function's value at that point.

• Removable Discontinuity: A function $f$ has a removable discontinuity at $x=c$ if the two-sided limit, ${\lim }_{x\to c}f(x)$, exists but is not equal to $f(c)$. This can happen if $f(c)$ is undefined or if $f(c)$ is defined as a different value than the limit.

  $$\underset{x\to c}{\lim }f(x)\text{ exists, but }\underset{x\to c}{\lim }f(x)\ne f(c)$$

• Jump Discontinuity: A function $f$ has a jump discontinuity at $x=c$ if the limit from the left, ${\lim }_{x\to c^{-}}f(x)$, and the limit from the right, ${\lim }_{x\to c^{+}}f(x)$, both exist as finite values but are not equal to each other.

  $$\underset{x\to c^{-}}{\lim }f(x)\ne \underset{x\to c^{+}}{\lim }f(x)(\text{where both limits are finite})$$

• Discontinuity due to a Vertical Asymptote: A function $f$ has a discontinuity due to a vertical asymptote at $x=c$ if either the limit from the left or the limit from the right (or both) is infinite.

  $$\underset{x\to c^{-}}{\lim }f(x)=\pm \infty \text{or}\underset{x\to c^{+}}{\lim }f(x)=\pm \infty$$

Understanding the Limit Conditions

The classification of a discontinuity is a direct consequence of limit analysis. The key is to systematically check the conditions for each type.

• The Two-Sided Limit is Key for Removable vs. Non-Removable: The first question to ask is: "Does ${\lim }_{x\to c}f(x)$ exist and is it finite?" This is equivalent to asking if the left-hand and right-hand limits are equal and finite. If the answer is yes, the discontinuity can only be removable (or the function is continuous if the limit equals $f(c)$). If the answer is no, the discontinuity must be either a jump or due to a vertical asymptote.

• Finite vs. Infinite Limits Distinguish Jumps from Asymptotes: If you have determined that the two-sided limit does not exist, the next step is to analyze why.

  • If ${\lim }_{x\to c^{-}}f(x)$ and ${\lim }_{x\to c^{+}}f(x)$ are both finite but different numbers, you have a jump discontinuity. The function approaches one concrete value from the left and a different concrete value from the right.

  • If at least one of the one-sided limits is infinite ($\pm \infty$), you have a discontinuity due to a vertical asymptote. The function's behavior is unbounded as it approaches $c$ from at least one side.

Core Concepts & Rules

• A discontinuity at $x=c$ is removable if the limits from the left and right side of $c$ are equal and finite, but this value does not equal $f(c)$.

• A discontinuity at $x=c$ is a jump discontinuity if the limits from the left and right side of $c$ are both finite but are not equal to each other.

• A discontinuity at $x=c$ is due to a vertical asymptote if the limit from either the left or the right side of $c$ (or both) is positive or negative infinity.

Step-by-Step Example 1: Analyzing a Piecewise Function

Problem: Given the function $f(x)$ defined below, classify the type of discontinuity at $x=1$.

$$f(x)=\{\begin{array}{ll}3x-1& x<1\\ 4& x=1\\ x^2+2& x>1\end{array}$$

Step 1: Evaluate the left-hand limit.

To find the limit as $x$ approaches 1 from the left, we use the piece of the function defined for $x<1$.

$$\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }(3x-1)=3(1)-1=2$$

Step 2: Evaluate the right-hand limit.

To find the limit as $x$ approaches 1 from the right, we use the piece of the function defined for $x>1$.

$$\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(x^2+2)=(1{)}^2+2=3$$

Step 3: Compare the one-sided limits.

The left-hand limit is 2, and the right-hand limit is 3. Both limits exist and are finite, but they are not equal.

$$\underset{x\to 1^{-}}{\lim }f(x)\ne \underset{x\to 1^{+}}{\lim }f(x)$$

Step 4: Classify the discontinuity.

According to the definition, since the one-sided limits exist as finite values but are not equal, the function $f(x)$ has a jump discontinuity at $x=1$.

Step-by-Step Example 2: Analyzing a Rational Function

Problem: Find and classify all discontinuities for the function $g(x)=\frac{x^2-9}{x^2+x-12}$.

Step 1: Factor the numerator and denominator.

Factoring helps identify common factors (potential removable discontinuities) and factors that cause the denominator to be zero (all potential discontinuities).

$$g(x)=\frac{(x-3)(x+3)}{(x+4)(x-3)}$$

Step 2: Identify potential points of discontinuity.

Discontinuities can only occur where the denominator of the original function is zero.

$x^2+x-12=0⟹(x+4)(x-3)=0$.

The potential discontinuities are at $x=3$ and $x=-4$.

Step 3: Analyze the discontinuity at $x=3$.

• Find the limit by first simplifying the expression for $x\ne 3$.

  $$\underset{x\to 3}{\lim }g(x)=\underset{x\to 3}{\lim }\frac{(x-3)(x+3)}{(x+4)(x-3)}=\underset{x\to 3}{\lim }\frac{x+3}{x+4}=\frac{3+3}{3+4}=\frac{6}{7}$$

• The two-sided limit exists and is equal to $\frac{6}{7}$.

• The function value $g(3)$ is undefined because it results in a $0/0$ form in the original function.

• Since ${\lim }_{x\to 3}g(x)$ exists but is not equal to $g(3)$ (as $g(3)$ is undefined), the function has a removable discontinuity at $x=3$.

Step 4: Analyze the discontinuity at $x=-4$.

• Find the limit using the simplified expression.

  $$\underset{x\to -4}{\lim }g(x)=\underset{x\to -4}{\lim }\frac{x+3}{x+4}$$

• As $x\to -4$, the numerator approaches $-1$ and the denominator approaches $0$. This indicates an infinite limit.

• To be thorough, check a one-sided limit:

  $$\underset{x\to -4^{+}}{\lim }\frac{x+3}{x+4}\to \frac{-1}{0^{+}}=-\infty$$

• Since at least one of the one-sided limits is infinite, the function has a discontinuity due to a vertical asymptote at $x=-4$.

Using Your Calculator

Classifying discontinuities is an analytical process based on evaluating limits. A graphing calculator should be used primarily to visualize and verify your analytical conclusions, not to find the answer directly.

To verify the analysis of $g(x)=\frac{x^2-9}{x^2+x-12}$ from Example 2:

1. Graph the Function: Enter Y_1 = (X^2-9)/(X^2+X-12). Use a standard zoom window (ZOOM 6).

   • You will see a vertical line near $x=-4$, which visually confirms the vertical asymptote.

   • The graph will appear to be smooth near $x=3$, but the calculator will not show the open circle for the removable discontinuity. This is a limitation of the display.

2. Use the Table to Approximate Limits:

   • Go to TBLSET (2nd + WINDOW) and change the Independent variable (Indpnt) to Ask.

   • Go to the TABLE (2nd + GRAPH).

   • To check $x=3$: Enter values very close to 3, like $2.999$ and $3.001$. The calculator will show Y-values very close to $0.85714$, which is $6/7$. This supports the conclusion that the limit is $6/7$. Entering $3$ itself will show an ERROR, confirming $g(3)$ is undefined. This evidence points to a removable discontinuity.

   • To check $x=-4$: Enter values very close to -4, like $-4.001$ (from the left) and $-3.999$ (from the right). The calculator will show a large positive value for $-4.001$ and a large negative value for $-3.999$. This supports the conclusion of a vertical asymptote with infinite one-sided limits.

AP Exam Quick Hit

Common Question Types

• Classifying from a Piecewise Function: Given a piecewise function, you will be asked to determine the type of discontinuity at a point where the definition changes. This requires evaluating left- and right-hand limits.

  • Example: "For $f(x)=\{\begin{array}{ll}\sin (\pi x)& x<2\\ (x-2{)}^2& x\ge 2\end{array}$, what type of discontinuity exists at $x=2$?"

• Classifying from a Rational Function: Given a rational function, you will be asked to find and classify all points of discontinuity. This requires factoring, canceling terms, and analyzing the remaining factors in the denominator.

  • Example: "Identify the location and type of each discontinuity for $h(x)=\frac{x-1}{x^2-1}$."

• Identifying from a Graph: You will be shown a graph of a function and asked to identify the x-values where specific types of discontinuities occur.

  • Example: "The graph of $f$ is shown above. At what value of $x$ does $f$ have a jump discontinuity?"

Common Mistakes

• Assuming all zeros in the denominator cause vertical asymptotes: A common error is to see a factor like $(x-c)$ in the denominator and immediately conclude there is a vertical asymptote at $x=c$. Students must check if the $(x-c)$ factor also exists in the numerator. If it cancels, the discontinuity is removable.

• Confusing "removable" with "hole": While a removable discontinuity is often visualized as a hole in the graph, the definition is precise: ${\lim }_{x\to c}f(x)\ne f(c)$. This includes cases where $f(c)$ is defined but at the wrong value, not just when it's undefined.

• Algebraic Errors in Factoring: Simple factoring mistakes when analyzing rational functions will lead to incorrect identification of the location and type of discontinuities. Always double-check your factoring.

• Using the wrong piece of a piecewise function: When evaluating ${\lim }_{x\to c^{-}}f(x)$, students must use the function rule for values less than$c$. A frequent mistake is to plug $c$ into the wrong part of the definition.