Problem 13 - Calculus for Biology and Medicine

Question

Show that the floor function \(f(x)=\lfloor x\rfloor\) is continuous at \(x=5 / 2\) but discontinuous at \(x=3\).

Step-by-Step Solution

Answer

The floor function is continuous at \( x = \frac{5}{2} \) and discontinuous at \( x = 3 \).

Step 1: Understand the Floor Function

The floor function, denoted as \( f(x) = \lfloor x \rfloor \), returns the greatest integer less than or equal to \( x \). For example, \( \lfloor 2.9 \rfloor = 2 \) and \( \lfloor 3 \rfloor = 3 \).

Step 2: Define Continuity at a Point

A function \( f(x) \) is continuous at a point \( x = a \) if \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \). This means the left-hand limit, right-hand limit, and the value of the function at \( x = a \) must all be equal.

Step 3: Evaluate Continuity at \( x = \frac{5}{2} \)

For \( x = \frac{5}{2} = 2.5 \), observe that \( f\left(\lfloor 2.5 \rfloor\right) = 2 \). The left-hand limit as \( x \to 2.5^- \) is 2 (since any number slightly less than 2.5 floors to 2), and the right-hand limit as \( x \to 2.5^+ \) is also 2 (since any number slightly more than 2.5 still floors to 2). Thus, \( \lim_{x \to 2.5^-} f(x) = \lim_{x \to 2.5^+} f(x) = f(2.5) = 2 \), showing continuity.

Step 4: Evaluate Discontinuity at \( x = 3 \)

For \( x = 3 \), observe that \( f(3) = 3 \). The left-hand limit as \( x \to 3^- \) is 2 (since any number slightly less than 3 floors to 2), but the right-hand limit as \( x \to 3^+ \) is 3 (since any number slightly more than 3 floors to 3). Therefore, \( \lim_{x \to 3^-} f(x) = 2 eq 3 = \lim_{x \to 3^+} f(x) = f(3) \), indicating a discontinuity.

Key Concepts

Continuity

Introduction to Continuity: Continuity in mathematics means that a function behaves predictably and without interruption at a given point. A function is continuous at a point if, as you approach that point from either side, the values of the function get closer and closer to the function's value at that exact point.

Understanding how this applies to the floor function is crucial. The floor function, denoted \( f(x) = \lfloor x \rfloor \), is indeed continuous at some points. To establish continuity at a particular point, say \( x = \frac{5}{2} \), we need to verify that the limits from the left side (approaching from below) and the right side (approaching from above) converge to the same value as the function's output at \( x = \frac{5}{2} \).

• For \( x = \frac{5}{2} \), the floor function gives us exactly 2 because the greatest integer less than or equal to 2.5 is 2.
• The value of the function, as well as the left and right-hand limits, all equal 2, proving continuity at this point.

Limits

Limits provide a method to examine the behavior of functions as they approach specific points. Simply put, a limit asks the question: As \( x \) gets very close to a specific value, what value does \( f(x) \) get close to?

For the floor function, viewing the limits helps construct understanding around where the function is continuous or not. Let's specifically consider the point \( x = \frac{5}{2} \):

• The left-hand limit (as \( x \to 2.5^- \)) refers to approaching 2.5 from values less than 2.5, which means values like 2.4999. The floor function turns all these values to \( 2 \).
• The right-hand limit (as \( x \to 2.5^+ \)) involves approaching 2.5 from above, like 2.5001, which are also floored to \( 2 \).

This uniformity around approaching a single common value illuminates why the floor function is continuous at this specific point. Limits unify the understanding by showing that approaching the function from both directions results in the same outcome.

Discontinuity

A function is said to be discontinuous at a point if there is any kind of jump or mismatch in the function's values around that point. Examining \( x = 3 \) for the floor function \( f(x) = \lfloor x \rfloor \), we find a characteristic discontinuity.

Here's how:

• When approaching 3 from the left, we consider numbers like 2.9999, which are floored to \( 2 \).
• Approaching from the right involves numbers like 3.0001, which the floor function rounds down to \( 3 \).
• At \( x = 3 \), the function value is \( 3 \).

The left-hand limit is 2, while the right-hand limit and the function's value at 3 is 3. Since the left-hand and right-hand limits are not equal, the function is discontinuous at this point. This jump from 2 to 3 without a smooth transition demonstrates why absolutely no continuity exists at \( x = 3 \).