Question

If \([\cdot]\) denotes the greatest integer function, then \(f(x) = \left[\frac{1}{2} - x\right] + [x]\)

• A. is continuous at \(x = \frac{1}{2}\)
• B. is discontinuous at \(x = \frac{1}{2}\)
• C. \(\lim_{x \to (1/2)^+} f(x) = 2\)
• D. \(\lim_{x \to (1/2)^-} f(x) = 1\)

Hint:

Greatest integer function has jump discontinuities at integer points.

Answer 1

The Correct Option is B

To determine whether the function \( f(x) = \left[\frac{1}{2} - x\right] + [x] \) is continuous at \( x = \frac{1}{2} \), we need to evaluate the left-hand limit, right-hand limit, and the function value at this point. The greatest integer function \([\cdot]\), also known as the floor function, returns the largest integer less than or equal to the given number.

1. Calculate the left-hand limit as \( x \to \frac{1}{2}^- \):
   For \( x \) slightly less than \( \frac{1}{2} \), \(\frac{1}{2} - x\) is slightly more than 0 (but less than 1), so \(\left[\frac{1}{2} - x\right] = 0\). Meanwhile, \([x]\) is 0 when \( x \) is slightly less than \( \frac{1}{2} \), so:
   \[\lim_{x \to (\frac{1}{2})^-} f(x) = 0 + 0 = 0\]
2. Calculate the right-hand limit as \( x \to \frac{1}{2}^+ \):
   For \( x \) slightly greater than \( \frac{1}{2} \), \(\frac{1}{2} - x\) is slightly less than 0, so \(\left[\frac{1}{2} - x\right] = -1\). Meanwhile, \([x]\) is still 0 when \( x \) is slightly greater than \( \frac{1}{2} \), so:
   \[\lim_{x \to (\frac{1}{2})^+} f(x) = -1 + 0 = -1\]
3. Evaluate \( f(x) \) at \( x = \frac{1}{2} \):
   At \( x = \frac{1}{2} \), \(\left[\frac{1}{2} - x\right] = 0\) and \([x] = 0\), hence:
   f(\frac{1}{2}) = 0 + 0 = 0
4. Check for continuity at \( x = \frac{1}{2} \):
   For \( f(x) \) to be continuous at \( x = \frac{1}{2} \), the left-hand limit, right-hand limit, and the function value at \(\frac{1}{2}\) must all be equal. Here, the left-hand limit is 0 and the right-hand limit is -1, showing they are not equal. Thus, \( f(x) \) is discontinuous at \( x = \frac{1}{2} \).

Therefore, the correct answer is is discontinuous at \( x = \frac{1}{2} \).