Question

If \( f(x) = \begin{cases} \frac{\sin |x|}{x}, & \text{for } [x] \ne 0 \\ 0, & \text{for } [x] = 0 \end{cases} \) where, \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\lim_{x \to 0} f(x)\) is equal to

• A. 1
• B. -1
• C. 0
• D. Does not exist

Hint:

For greatest integer function, always check left-hand and right-hand limits separately.

Answer 1

The Correct Option is D

Concept: Behavior of function depends on greatest integer function \([x]\) near \(x=0\).

Step 1: Consider right-hand limit \(x \to 0^+\).
For \(0<x<1\), we have: \[ [x] = 0 \Rightarrow f(x) = 0 \] Thus, \[ \lim_{x \to 0^+} f(x) = 0 \]

Step 2: Consider left-hand limit \(x \to 0^-\).
For \(-1<x<0\), we have: \[ [x] = -1 \Rightarrow f(x) = \frac{\sin |x|}{x} \] Since \(|x| = -x\) (for negative \(x\)): \[ f(x) = \frac{\sin(-x)}{x} = \frac{-\sin x}{x} \] \[ \lim_{x \to 0^-} f(x) = -1 \]

Step 3: Compare limits.
\[ \lim_{x \to 0^+} f(x) = 0,\quad \lim_{x \to 0^-} f(x) = -1 \] Since LHL \(\neq\) RHL, limit does not exist.