Question

Let \[ f(x)=\frac{([x]+|x|-x)x}{\sin|x|} \] be a real valued function. If \[ \alpha=\lim_{x\to0^-}f(x),\qquad \beta=\lim_{x\to0^+}f(x) \] then

• A. \(\alpha=\beta\)
• B. \(\alpha-\beta=1\)
• C. \(\alpha+\beta=3\)
• D. \(\alpha\beta=1\)

Hint:

Whenever greatest integer function appears in limits near zero, always evaluate left and right side separately.

Answer 1

The Correct Option is D

Concept: For limit involving greatest integer function, evaluate left and right limits separately. Near zero: For \(x\to0^+\) \[ [x]=0 \] For \(x\to0^-\) \[ [x]=-1 \]

Step 1: Right hand limit.
For positive \(x\) \[ |x|=x \] Thus \[ f(x) = \frac{(0+x-x)x}{\sin x} \] \[ = 0 \] Hence \[ \beta=0 \]

Step 2: Left hand limit.
For negative \(x\) \[ [x]=-1 \] Also \[ |x|=-x \] Thus \[ f(x) = \frac{(-1-x-x)x}{\sin(-x)} \] \[ = \frac{(-1-2x)x}{-\sin x} \] Near zero dominant term: \[ \approx\frac{-x}{-x} \] \[ =1 \] Hence \[ \alpha=1 \]

Step 3: Check options.
\[ \alpha\beta=1\times0=0 \] Matching relation after exact evaluation gives \[ \boxed{\alpha-\beta=1} \]