Problem 55 - Calculus: Early Transcendentals

Question

If \( f(x) = [ x ] + [ -x ] \), show that \( \displaystyle \lim_{x \to 2}f(x) \) exists but is not equal to \(f(2) \).

Step-by-Step Solution

Answer

The limit exists and equals -1, but \( f(2) = 0 \) so they are not equal.

Step 1: Understanding the Function

The function is defined as \( f(x) = [x] + [-x] \) where \([x]\) represents the greatest integer less than or equal to \(x\), known as the floor function. Our task is to evaluate its behavior around \(x = 2\).

Step 2: Calculate \( f(2) \)

By substituting \(2\) into the expression, \([2] + [-2] = 2 - 2 = 0\). Therefore, \( f(2) = 0 \).

Step 3: Evaluate \( \\lim_{x \to 2^+} f(x) \)

For \(x\) slightly greater than 2, such as \(x = 2.1\), \([x] = 2\) and \([-x] = -3\). Hence, \([x] + [-x] = 2 - 3 = -1\). Therefore, as \(x\) approaches 2 from the right, \(f(x) = -1.\)

Step 4: Evaluate \( \\lim_{x \to 2^-} f(x) \)

For \(x\) slightly less than 2, such as \(x = 1.9\), \([x] = 1\) and \([-x] = -2\). Hence, \([x] + [-x] = 1 - 2 = -1\). Therefore, as \(x\) approaches 2 from the left, \(f(x) = -1\).

Step 5: Evaluate the Two-Sided Limit

Since both the left-hand limit and the right-hand limit as \(x\) approaches 2 are equal, \( \lim_{x \to 2} f(x) = -1. \)

Step 6: Conclusion

The two-sided limit at \(x = 2\) exists and equals \(-1\). However, \(f(2) = 0\), so \( \lim_{x \to 2} f(x) eq f(2) \).

Key Concepts

Floor Function

The floor function, denoted by \([x]\), is an important concept in mathematics. It refers to the greatest integer less than or equal to a specific value. For example, if you have a number like 2.9, the floor function \([2.9]\) would be 2, because 2 is the largest integer not exceeding 2.9.
It's a simple yet powerful tool for rounding numbers down to the nearest whole number.

• If the number is already an integer, like 3, then \([3] = 3\).
• If the number has a decimal, like 4.7, then \([4.7] = 4\).

Understanding how the floor function interacts with positive and negative numbers is critical. When used on negative numbers, such as -2.3, the result \([-2.3]\) is -3, as it's the largest integer less than -2.3. This understanding helps when evaluating functions like \(f(x) = [x] + [-x]\), especially when assessing limits.

Two-Sided Limit

A two-sided limit is an essential concept in calculus that examines what value a function approaches as the input approaches a certain point from both sides. This requires evaluating both the left-hand and right-hand limits and ensuring they match.
For a function \(f(x)\) to have a two-sided limit at \(x = a\), both these conditions must hold:

• \( \lim_{x \to a^-} f(x) \) exists
  
• \( \lim_{x \to a^+} f(x) \) exists
  
• They must be equal, i.e., \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)\)

In the context of \(f(x) = [x] + [-x]\), the two-sided limit as \(x \to 2\) was calculated by considering the behavior of the function from values slightly less than 2 and slightly more than 2. Both directional limits approached -1, establishing that \lim_{x \to 2}f(x) = -1\.

Left-Hand and Right-Hand Limits

To properly understand the behavior of a function at a given point, it's crucial to consider both left-hand and right-hand limits. Left-hand limit refers to the value that a function approaches as the variable approaches a particular number from the left, denoted as \(x \to a^-\), while the right-hand limit considers approach from the right, \(x \to a^+\).
In examining \(f(x) = [x] + [-x]\) at \(x = 2\), we find:

• For the left-hand limit \((x \to 2^-)\), substituting values like 1.9, yields the integer parts \([1.9] + [-1.9] = 1 + (-2) = -1\).
• For the right-hand limit \((x \to 2^+)\), using values like 2.1 gives \([2.1] + [-2.1] = 2 + (-3) = -1\).

As both limits equal -1, the function showcases consistent behavior from both directions, supporting the two-sided limit existence.

Piecewise Functions

Piecewise functions define various behaviors for different intervals of the input values, creating distinct "pieces" of the overall function. In the case of \(f(x) = [x] + [-x]\), although it is not explicitly presented as a piecewise function, understanding its behavior can relate it closely to this concept.
When you consider intervals over which \([x]\) and \([-x]\) might vary, it produces segments where the function seems to operate differently. Such functions often involve using limits to analyze their behavior at boundaries, where rule changes occur, like around non-integers transitioning between floor values.
When examining the function around x = 2, knowing that \(f(2) = 0\) and \( \lim_{x \to 2} f(x) = -1\) highlights how localized behaviors in functions can lead to differing outputs and limits. The treatment in different mathematical intervals and conditions is akin to analyzing piecewise function behaviors.