Problem 57 - Thomas Calculus

Question

Removable discontinuity Give an example of a function \(f(x)\) that is continuous for all values of \(x\) except \(x=2,\) where it has a removable discontinuity. Explain how you know that \(f\) is discontinuous at \(x=2,\) and how you know the discontinuity is removable.

Step-by-Step Solution

Answer

The function \( f(x) = \frac{(x-2)(x+3)}{x-2} \) has a removable discontinuity at \( x=2 \).

Step 1: Define the Function

Let's consider a function \( f(x) = \frac{(x-2)(x+3)}{x-2} \). This function is continuous except at \( x = 2 \) because the denominator becomes zero, leading to an undefined point.

Step 2: Evaluate the Limit

To determine if the discontinuity at \( x = 2 \) is removable, evaluate the limit of \( f(x) \) as \( x \) approaches 2:\[ \lim_{{x \to 2}} \frac{(x-2)(x+3)}{x-2} = \lim_{{x \to 2}} (x+3) = 5. \]This shows that the limit exists and is equal to 5.

Step 3: Check Continuity Without the Factor

To check for removable discontinuity, simplify \( f(x) \) by canceling out the \( x-2 \) factor:\[ f(x) = x+3 \text{ for } x eq 2. \]This new function, \( g(x) = x+3 \), is continuous everywhere including \( x = 2 \).

Step 4: Conclude Removable Discontinuity

The original function \( f(x) \) had a discontinuity only at \( x=2 \), but \( f(x) \) approaches a finite limit at this point (5) and can be made continuous by redefining \( f(2) = 5 \). Therefore, the discontinuity at \( x=2 \) is removable.

Key Concepts

Limit Calculation

When dealing with removable discontinuities, one of the first steps is to analyze the limits. In our example, we need to find the limit of the function \( f(x) = \frac{(x-2)(x+3)}{x-2} \) as \( x \) approaches 2. Because little confusions around limits often occur, let's dive into how these calculations work.
Calculating the limit involves understanding how the function behaves as \( x \) gets infinitely close to 2, but never actually reaching it. In cases like this one, you must simplify the expression first by canceling common factors in the numerator and denominator. Once simplified, our function is just \( x+3 \).
Next, substitute 2 into the simplified function to find \( \lim_{{x \to 2}} (x+3) \). By substituting, you get \( 2+3 \), which is equal to 5. This tells us the function approaches a value of 5 as \( x \) gets closer to 2 from either side.

Continuity

In calculus, continuity focuses on whether a function behaves well at all points of its domain. Simply put, a function is continuous at a point \( x = a \) if \( f(a) \) is defined, the limit as \( x \) approaches \( a \) exists, and these two values are equal.
For the function \( f(x) = \frac{(x-2)(x+3)}{x-2} \), although the limit as \( x \to 2 \) exists (5 in our case), \( f(2) \) is undefined because of the division by zero. Thus, \( f \) has a discontinuity at \( x = 2 \).
This type of discontinuity is termed "removable" because we can redefine \( f(2) \) to make the function continuous at this specific point. By setting \( f(2) = 5 \), all three conditions for continuity are satisfied, thus transforming the original function into one that is completely continuous across its domain.

Function Simplification

Simplifying functions is a critical part of understanding where discontinuities like the removable type occur. The goal is to turn a complex function into a simpler form without changing its behavior. In our problem, we simplified \( f(x) \) by canceling out the common \( x-2 \) factor in both the numerator and the denominator.
Once simplified, the function becomes \( g(x) = x + 3 \), which is much easier to manipulate and analyze. This simpler form helps us in assessing limit and continuity since it clearly shows the function's behavior across its entire domain except at the removable discontinuity point.
The act of simplifying is powerful in detecting and resolving discontinuities. Especially removable ones, which can often be fixed, making the function whole and continuous again by adjusting just a single point.