Question

Given: \[ \lim_{x \to a} \left\lfloor \left( \frac{x^3}{a} - \left\lfloor \frac{x}{a} \right\rfloor^3 \right) \right\rfloor = k, \quad \lim_{x \to a} \left( \left\lfloor \frac{x^3}{a} \right\rfloor - \left\lfloor \frac{x}{a} \right\rfloor^3 \right) = l \] Then: \[ \boxed{?} \]

• A. \( k = l \)
• B. \( k - l = 1 \)
• C. \( |l - k| = 1 \)
• D. \( l = a^2, k \text{ does not exist} \)

Hint:

Limits involving greatest integer functions often differ subtly based on the order of operations — evaluate each direction near the point carefully.

Answer 1

The Correct Option is B

Let \( x \to a^- \), then:
- \( \left\lfloor \frac{x}{a} \right\rfloor = 0 \)
- \( \left\lfloor \frac{x^3}{a} \right\rfloor = 0 \)
- So: \( \left\lfloor \frac{x^3}{a} - \left\lfloor \frac{x}{a} \right\rfloor^3 \right\rfloor = \left\lfloor \frac{x^3}{a} \right\rfloor = 0 \)
Now take \( x \to a^+ \), then: - \( \left\lfloor \frac{x}{a} \right\rfloor = 1 \)
- \( \left\lfloor \frac{x^3}{a} \right\rfloor = 1 \)
- \( \left\lfloor \frac{x^3}{a} - 1 \right\rfloor = 0 \)
Hence: - First limit (inside floor first): \( k = 1 \)
- Second limit (floor first then subtract): \( l = 0 \)
So: \[ k - l = 1 \]