Removing Discontinuities - AP Calculus BC Study Guide

The Core Idea: Removing Discontinuities

A function is continuous at a point if its graph can be drawn through that point without lifting your pencil. A discontinuity is a break in the graph. This topic focuses on a specific type of break called a "removable discontinuity." This occurs when a function's limit exists at a point, $x=c$, but the function itself is either not defined at $c$ or has a value different from the limit. The core idea is that this type of discontinuity can be "repaired" or "removed."

We can create a new, continuous function by defining or redefining the function's value at that single point to be equal to the value of the limit. This process effectively patches the hole in the function's graph, making it continuous at that specific point without altering the function's behavior elsewhere. The entire process hinges on the crucial condition that the limit of the function at the point of discontinuity must exist.

Key Definitions

The concept of removing a discontinuity is built upon the definition of a removable discontinuity, which occurs in two specific situations.

1. Removable Discontinuity (Function Undefined)

A function $f$ has a removable discontinuity at $x=c$ if:

• $f(c)$ is not defined.

• ${\lim }_{x\to c}f(x)$ exists and is equal to a finite value $L$.

To remove this discontinuity, we define a new function, often called an extension of $f$, which is continuous at $x=c$. This new function, let's call it $g(x)$, is defined such that $g(c)={\lim }_{x\to c}f(x)$.

2. Removable Discontinuity (Function Value Mismatches Limit)

A function $f$ has a removable discontinuity at $x=c$ if:

• $f(c)$ is defined.

• ${\lim }_{x\to c}f(x)$ exists and is equal to a finite value $L$.

• However, $f(c)\ne L$.

To remove this discontinuity, we redefine the function's value at $x=c$. The new, continuous function $g(x)$ is defined such that $g(c)={\lim }_{x\to c}f(x)=L$.

In both cases, the action of "removing" the discontinuity is to ensure the function's value at the point equals the function's limit at that point.

Understanding the Conditions for Removal

The ability to remove a discontinuity is entirely dependent on one critical condition: the limit of the function must exist at the point of discontinuity. This is the foundational requirement from which everything else follows.

Recall the three conditions for a function $f$ to be continuous at a point $x=c$:

1. $f(c)$ must be defined.

2. ${\lim }_{x\to c}f(x)$ must exist.

3. ${\lim }_{x\to c}f(x)=f(c)$.

A removable discontinuity occurs precisely when Condition 2 is met, but either Condition 1 or Condition 3 fails.

• If $f(c)$ is undefined, Condition 1 fails. If the limit exists, we can define $f(c)$ to be the value of the limit, thereby satisfying all three conditions.

• If $f(c)$ is defined but not equal to the limit, Condition 3 fails. If the limit exists, we can redefine $f(c)$ to be the value of the limit, again satisfying all three conditions.

If the limit as $x\to c$ does not exist (for example, in the case of a jump or an infinite discontinuity), the discontinuity is non-removable. No matter what value we assign to $f(c)$, we can never satisfy the third condition of continuity, ${\lim }_{x\to c}f(x)=f(c)$, because the limit itself does not exist.

Core Concepts & Rules

• A discontinuity at a point $x=c$ is classified as removable if and only if the limit, ${\lim }_{x\to c}f(x)$, exists as a finite number.

• A removable discontinuity occurs under two possible circumstances:

  1. The function is not defined at $x=c$, but the limit as $x\to c$ exists.

  2. The function is defined at $x=c$, but its value, $f(c)$, is not equal to the limit as $x\to c$.

• To remove the discontinuity at $x=c$, one must define or redefine the function's value at that point to be equal to the limit.

• This process creates a new function, $g(x)$, that is continuous at $x=c$. This new function is typically defined as a piecewise function:

  $$g(x)=\{\begin{array}{ll}f(x)& x\ne c\\ {\lim }_{x\to c}f(x)& x=c\end{array}$$

Step-by-Step Example 1: Function Undefined at a Point

Problem: Consider the function $f(x)=\frac{x^2+x-6}{x-2}$. Show that $f$ has a removable discontinuity at $x=2$ and define an extension of $f$, let's call it $g(x)$, that is continuous at $x=2$.

Step 1: Identify the point of discontinuity.

The function $f(x)$ is a rational function. Its denominator is zero when $x-2=0$, which means $x=2$. Therefore, $f(2)$ is undefined, and the function is discontinuous at $x=2$.

Step 2: Evaluate the limit at the point of discontinuity.

To determine if the discontinuity is removable, we must check if the limit exists. We can do this by factoring the numerator.

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

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

$$\underset{x\to 2}{\lim }(x+3)=2+3=5$$

Step 3: Conclude that the discontinuity is removable.

The limit as $x\to 2$ exists and is equal to 5, but $f(2)$ is undefined. This matches the definition of a removable discontinuity.

Step 4: Define the new, continuous function.

To remove the discontinuity, we define a new function $g(x)$ that is identical to $f(x)$ for all $x\ne 2$, but we define its value at $x=2$ to be the limit we found.

$$g(x)=\{\begin{array}{ll}\frac{x^2+x-6}{x-2}& x\ne 2\\ 5& x=2\end{array}$$

This function $g(x)$ is now continuous at $x=2$.

Step-by-Step Example 2: Function Value Mismatches Limit

Problem: Let $f(x)$ be the piecewise function defined below. Determine if $f(x)$ has a removable discontinuity at $x=0$. If so, redefine $f(x)$ to make it continuous at $x=0$.

$$f(x)=\{\begin{array}{ll}\frac{e^{2x}-1}{x}& x\ne 0\\ 1& x=0\end{array}$$

Step 1: Identify the function's value at the point.

The function is explicitly defined at $x=0$. According to the definition, $f(0)=1$.

Step 2: Evaluate the limit at the point of discontinuity.

We need to find ${\lim }_{x\to 0}f(x)$. For $x\ne 0$, $f(x)=\frac{e^{2x}-1}{x}$.

$$\underset{x\to 0}{\lim }\frac{e^{2x}-1}{x}$$

Direct substitution yields $\frac{e^0-1}{0}=\frac{1-1}{0}=\frac{0}{0}$, which is an indeterminate form. We can use L'Hôpital's Rule.

$$\underset{x\to 0}{\lim }\frac{\frac{d}{dx}(e^{2x}-1)}{\frac{d}{dx}(x)}=\underset{x\to 0}{\lim }\frac{2e^{2x}}{1}$$

Now, we can use direct substitution:

$$\frac{2e^{2(0)}}{1}=\frac{2(1)}{1}=2$$

The limit of $f(x)$ as $x\to 0$ is 2.

Step 3: Compare the limit to the function's value.

We found that ${\lim }_{x\to 0}f(x)=2$, but the function is defined as $f(0)=1$. Since the limit exists but is not equal to the function's value ($2\ne 1$), $f(x)$ has a removable discontinuity at $x=0$.

Step 4: Redefine the function to make it continuous.

To remove the discontinuity, we must change the value of $f(0)$ to match the limit. The new, continuous function, $g(x)$, is:

$$g(x)=\{\begin{array}{ll}\frac{e^{2x}-1}{x}& x\ne 0\\ 2& x=0\end{array}$$

Using Your Calculator

The process of identifying and removing a discontinuity is fundamentally analytical. You must use algebraic techniques and limit laws to find the limit and redefine the function. A calculator cannot perform these steps for you.

However, a graphing calculator can be a powerful tool for visualizing the problem and verifying your analytical answer.

1. Visualizing the Discontinuity:

• Graph the function in question. For $f(x)=\frac{x^2+x-6}{x-2}$ from Example 1, the graph will look like the line $y=x+3$.

• The calculator screen may not show the "hole" at $x=2$. You can use the TRACE feature and type in $x=2$. The calculator will show no $y$-value, confirming that the function is undefined there.

2. Approximating the Limit:

• Use the table feature of your calculator (2nd + GRAPH).

• Set the table to "Ask" for the independent variable (TBLSET -> $Indpnt:Ask$).

• Enter values very close to the point of discontinuity from both the left and the right. For $x=2$, you could enter $1.9$, $1.99$, $1.999$ and $2.1$, $2.01$, $2.001$.

• Observe the corresponding $y$-values. You will see them approaching the limit you calculated analytically (in this case, 5). This provides strong evidence that your limit calculation is correct.

AP Exam Quick Hit

Common Question Types

• Finding a Constant to Ensure Continuity: You will be given a piecewise function with an unknown constant, $k$, and asked to find the value of $k$ that makes the function continuous.

  • Example: "Let $f(x)=\{\begin{array}{ll}\frac{\sin (x)}{5x}& x\ne 0\\ k& x=0\end{array}$. For what value of $k$ is $f$ continuous at $x=0$?" (You would find ${\lim }_{x\to 0}\frac{\sin (x)}{5x}=\frac{1}{5}$ and set $k=\frac{1}{5}$.)

• Defining a Continuous Extension: You will be given a function with a removable discontinuity and asked to write the definition of a new function that is continuous at that point.

  • Example: "Let $f(x)=\frac{x-4}{\sqrt{x}-2}$. The function $f$ has a removable discontinuity at $x=4$. Write an extended function $g$ that is continuous at $x=4$." (You would find the limit, which is 4, and write the piecewise definition for $g$.)

Common Mistakes

• Incorrectly Applying L'Hôpital's Rule: Students may try to use L'Hôpital's Rule when the limit is not in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty }{\infty }$), or they may make errors when taking the derivatives of the numerator or denominator.

• Algebraic Errors: Simple mistakes in factoring, simplifying complex fractions, or multiplying by a conjugate will lead to an incorrect limit value. For example, incorrectly factoring $x^2+x-6$ as $(x+2)(x-3)$.

• Stopping at $0/0$: When direct substitution results in $\frac{0}{0}$, some students mistakenly conclude that the limit does not exist. This is incorrect; $\frac{0}{0}$ is an indeterminate form that signals more work is needed (e.g., factoring, L'Hôpital's Rule) to find the limit.

• Confusing Removable with Non-Removable Discontinuities: A student might find that ${\lim }_{x\to c}f(x)$ does not exist (e.g., it's a jump discontinuity) but still try to "remove" it by setting $f(c)$ to a one-sided limit. Remember, if the two-sided limit does not exist, the discontinuity is non-removable.