Removing Discontinuities - AP Calculus AB Study Guide

The Core Idea: Removing Discontinuencies

A function is continuous at a point if its graph can be drawn through that point without lifting your pencil. This intuitive idea is formalized by three conditions: the function must have a defined value at the point, a limit must exist as we approach the point, and these two values must be equal. Sometimes, a function fails this test in a very specific way: the limit exists, but the function value either doesn't exist or doesn't match the limit. This creates a "hole" in the graph, known as a removable discontinuity.

The core idea of this topic is that such a discontinuity can be "removed" or "repaired." If we can determine the value the function is approaching (the limit), we can create a new, related function that is continuous at that point. We do this by defining or redefining the function's value at the point of discontinuity to be equal to the value of the limit. This effectively "plugs the hole" and makes the function continuous at that specific point.

Key Definitions

The concept of removing a discontinuity is built upon the formal definition of continuity and the specific characteristics of a removable discontinuity.

The Three Conditions for Continuity at a Point

A function $f(x)$ is continuous at a point $x=c$ if and only if all three of the following conditions are met:

1. $f(c)$ exists (The function has a defined value at $c$).

2. ${\lim }_{x\to c}f(x)$ exists (The function approaches a single finite value as $x$ approaches $c$ from both sides).

3. ${\lim }_{x\to c}f(x)=f(c)$ (The limit at $c$ is equal to the function's value at $c$).

Removable Discontinuity

A removable discontinuity exists at $x=c$ if the limit of the function exists at $c$, but the function is not continuous there. This occurs when one of the other conditions for continuity fails:

• ${\lim }_{x\to c}f(x)$ exists, but $f(c)$ does not exist.

• ${\lim }_{x\to c}f(x)$ exists and $f(c)$ exists, but ${\lim }_{x\to c}f(x)\ne f(c)$.

Visually, this corresponds to a hole in the graph of the function.

Understanding How to Remove a Discontinuity

The term "removable" is used because we can eliminate the discontinuity by defining or redefining the function at the single point $x=c$. The key is that the limit must exist. The existence of the limit tells us exactly where the "hole" in the graph is located in the y-direction. It provides the specific value needed to plug the hole and make the function continuous.

The process involves creating a new function, often a piecewise-defined function, that is identical to the original function everywhere except at the point of discontinuity. At that specific point, the new function's value is explicitly defined to be the value of the limit.

If a function $f(x)$ has a removable discontinuity at $x=c$, we can define a new function $g(x)$ that is continuous at $c$:

$$g(x)=\{\begin{array}{llc}f(x),& x\ne c\text{n}{\lim }_{x\to c}f(x),& x=c\end{array}$$

By setting $g(c)={\lim }_{x\to c}f(x)$, we ensure that all three conditions for continuity are met for the function $g(x)$ at $x=c$.

Core Concepts & Rules

• Continuity Conditions: For a function $f$ to be continuous at $x=c$, three things must be true: $f(c)$ is defined, ${\lim }_{x\to c}f(x)$ exists, and $f(c)={\lim }_{x\to c}f(x)$.

• Identifying Removable Discontinuities: A removable discontinuity occurs at $x=c$ when ${\lim }_{x\to c}f(x)$ exists, but the function is still discontinuous at $c$. This is typically because $f(c)$ is undefined or $f(c)$ is not equal to the limit. These are graphically represented as holes.

• The Removal Process: A removable discontinuity at $x=c$ can be fixed. The fix is to define (or redefine) the function's value at $c$ to be equal to the limit as $x$ approaches $c$.

Step-by-Step Example 1: Identifying and Removing a Discontinuity

Consider the function $f(x)=\frac{x^2-5x+6}{x-2}$. Determine if the function has a removable discontinuity, and if so, create a new function $g(x)$ that is continuous everywhere.

Step 1: Check for points of discontinuity.

The function is a rational function, so it is discontinuous where the denominator is zero.

$x-2=0⟹x=2$.

The function is discontinuous at $x=2$.

Step 2: Check if the discontinuity is removable by evaluating the limit.

To see if the discontinuity is removable, we must check if the limit exists at $x=2$. We can do this by factoring the numerator.

$$\underset{x\to 2}{\lim }\frac{x^2-5x+6}{x-2}=\underset{x\to 2}{\lim }\frac{(x-2)(x-3)}{x-2}$$

Since we are taking the limit as $x$approaches 2, $x\ne 2$, so we can cancel the $(x-2)$ terms.

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

Because the limit exists ($-1$), but the function value $f(2)$ is undefined, the function has a removable discontinuity at $x=2$. The hole in the graph is at the point $(2,-1)$.

Step 3: Define a new function that removes the discontinuity.

We can remove the discontinuity by defining a new function $g(x)$ that is equal to $f(x)$ for all $x\ne 2$, and is equal to the limit at $x=2$.

$$g(x)=\{\begin{array}{llc}\frac{x^2-5x+6}{x-2},& x\ne 2\text{n}-1,& x=2\end{array}$$

This function $g(x)$ is now continuous at $x=2$ because $g(2)=-1$ and ${\lim }_{x\to 2}g(x)=-1$.

Step-by-Step Example 2: Finding a Parameter for Continuity

Let the function $f(x)$ be defined as:

$$f(x)=\{\begin{array}{llc}\frac{\sin (4x)}{x},& x\ne 0\text{nk},& x=0\end{array}$$

Find the value of the constant $k$ that makes the function $f(x)$ continuous at $x=0$.

Step 1: Identify the conditions for continuity at $x=0$

For $f(x)$ to be continuous at $x=0$, we must satisfy the condition ${\lim }_{x\to 0}f(x)=f(0)$.

Step 2: Evaluate the function value at the point.

From the definition of the piecewise function, we know that $f(0)=k$.

Step 3: Evaluate the limit as $x$ approaches the point.

We need to find the limit of the first piece of the function as $x\to 0$.

$$\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{\sin (4x)}{x}$$

This limit is related to the known trigonometric limit ${\lim }_{\theta \to 0}\frac{\sin (\theta )}{\theta }=1$. We can use algebraic manipulation to match this form.

$$\underset{x\to 0}{\lim }\frac{\sin (4x)}{x}=\underset{x\to 0}{\lim }4\cdot \frac{\sin (4x)}{4x}$$

Let $\theta =4x$. As $x\to 0$, $\theta \to 0$.

$$4\cdot \underset{\theta \to 0}{\lim }\frac{\sin (\theta )}{\theta }=4\cdot 1=4$$

So, ${\lim }_{x\to 0}f(x)=4$.

Step 4: Set the limit equal to the function value to find $k$

To ensure continuity, we set the results from Step 2 and Step 3 equal to each other.

$f(0)={\lim }_{x\to 0}f(x)$

$k=4$

Therefore, the value $k=4$ makes the function $f(x)$ continuous at $x=0$.

Using Your Calculator

This topic is primarily analytical, meaning problems are solved by hand using algebraic techniques and limit rules. A graphing calculator is not used to find the value that removes a discontinuity, but it can be an excellent tool for visualizing the problem and checking your answer.

To check the result of Example 1:

1. Press Y= and enter the function $Y_1=(X^2-5X+6)/(X-2)$.

2. Press GRAPH. You may need to use $ZOOM:6.ZStandard$. The graph will look like a straight line, but the calculator does not explicitly show the hole.

3. To verify the hole, press 2nd then $CALC$ (TRACE) and select $1:value$.

4. When prompted for $X=$, enter $2$ and press ENTER. The calculator will return a blank Y= value, confirming that the function is undefined at $x=2$.

5. To check the limit, you can use the table feature. Press 2nd then TABLE (GRAPH). Set your table to start near 2$with a small step size (e.g., `TblStart=1.99`, `ΔTbl=0.01`). You will see that as $X gets closer to 2, the Y1 values get closer to$-1$, confirming your calculated limit.

AP Exam Quick Hit

Common Question Types

• Finding a parameter for continuity: You will be given a piecewise function with a constant, like $k$ or $c$, and asked to find the value of the constant that makes the function continuous at the point where the function definition changes.

  • Example: Find the value of $c$ that makes $f(x)=\{\begin{array}{ll}\frac{x^2-16}{x-4},& x\ne 4\\ c,& x=4\end{array}$ continuous at $x=4$.

• Identifying the location of a removable discontinuity: You will be given a rational function and asked to find the coordinates $(x,y)$ of its removable discontinuity (hole). This requires finding the x-value that makes the denominator zero and then finding the limit at that x-value to determine the y-coordinate.

  • Example: Find the coordinates of the removable discontinuity of the function $g(x)=\frac{x+1}{x^2-x-2}$.

Common Mistakes

• Confusing Removable and Non-Removable Discontinuities: A common error is to misidentify a vertical asymptote as a removable discontinuity. Remember, for a discontinuity at $x=c$ to be removable, the limit ${\lim }_{x\to c}f(x)$ must be a finite number. If the limit is $\infty$ or $-\infty$, it is a non-removable infinite discontinuity (a vertical asymptote).

• Only Finding the x-coordinate: When asked for the location of a removable discontinuity, students often correctly find the x-value by setting the canceled factor to zero but forget to find the corresponding y-value by calculating the limit.

• Algebraic Errors: Simple mistakes in factoring polynomials or simplifying complex fractions can lead to an incorrect limit value, and therefore an incorrect value for the parameter or y-coordinate of the hole.

• Incorrectly Applying Limit Rules: Forgetting special trigonometric limits like ${\lim }_{x\to 0}\frac{\sin x}{x}=1$ or misapplying them in different forms can lead to errors in problems like Example 2.