Question

Let \( \left\lfloor x \right\rfloor \) be the greatest integer less than or equal to \( x \). Then \[ \lim_{x \to 0^-} \frac{x \left( \left\lfloor x \right\rfloor + |x| \right)}{|x|} \] is equal to:

• A. -1
• B. -2
• C. 0
• D. 1
• E. 2

Hint:

When dealing with floor functions in limits, carefully examine the behavior of \( \left\lfloor x \right\rfloor \) as \( x \) approaches the desired point. For negative values of \( x \), \( \left\lfloor x \right\rfloor \) is the greatest integer less than or equal to \( x \), and \( |x| = -x \).

Answer 1

The Correct Option is D

We are asked to evaluate the limit: \[ L = \lim_{x \to 0^-} \frac{x \left( \left\lfloor x \right\rfloor + |x| \right)}{|x|} \] Step 1: Consider the behavior of \( \left\lfloor x \right\rfloor \) and \( |x| \) as \( x \to 0^- \):
- As \( x \) approaches 0 from the left, \( x \) is negative, so \( |x| = -x \).
- For \( x \in (-1, 0) \), \( \left\lfloor x \right\rfloor = -1 \), since \( \left\lfloor x \right\rfloor \) is the greatest integer less than or equal to \( x \).
Step 2: Substitute these values into the expression: \[ L = \lim_{x \to 0^-} \frac{x \left( -1 + (-x) \right)}{-x} \] Simplify the expression: \[ L = \lim_{x \to 0^-} \frac{x \left( -1 - x \right)}{-x} = \lim_{x \to 0^-} \frac{x(-1 - x)}{-x} \] \[ L = \lim_{x \to 0^-} \left( 1 + x \right) \] Step 3: Now, take the limit as \( x \to 0^- \): \[ L = 1 + 0 = 1 \]
Thus, the value of the limit is \( 1 \).
Therefore, the correct answer is option (D).