Question

Let [t] denote the greatest integer \(\le\) t. If for some \(\lambda \in \mathbb{R} - \{1, 0\}\),
\(\lim_{x \to 0}\left|\frac{1 - x + |x|}{\lambda - x + [x]}\right| = L\), then L is equal to :

• A. 1
• B. 2
• C. 0
• D. \(\frac{1}{2}\)

Hint:

Evaluate the left and right limits separately (where [x] = -1 and 0 respectively); existence of the limit forces \(|\lambda| = |\lambda - 1|\).

Answer 1

The Correct Option is B

Right-hand limit (\(x \to 0^+\)): \(|x| = x\), \([x] = 0\), so the expression is \(\left|\frac{1}{\lambda - x}\right| \to \frac{1}{|\lambda|}\).
Left-hand limit (\(x \to 0^-\)): \(|x| = -x\), \([x] = -1\), so the expression is \(\left|\frac{1 - 2x}{\lambda - x - 1}\right| \to \frac{1}{|\lambda - 1|}\).
The limit L exists, so \(|\lambda| = |\lambda - 1|\), giving \(\lambda = \frac{1}{2}\).
Then \(L = \frac{1}{|1/2|} = 2\).
Answer: (2) 2