JEE PYQ: Limits Question 11
Question 11 - 2020 (03 Sep Shift 1)
Let $[t]$ denote the greatest integer $\leq t$. If for some $\lambda \in \mathbb{R} - {0, 1}$, $\lim_{x \to 0} \left|\frac{1 - x + |x|}{\lambda - x + [x]}\right| = L$, then $L$ is equal to:
(1) $1$
(2) $2$
(3) $\frac{1}{2}$
(4) $0$
Show Answer
Answer: (2)
Solution
For LHL: $\lim_{h\to 0} \left|\frac{1+h+h}{\lambda+h-1}\right| = \frac{1}{|\lambda-1|}$. For RHL: $\lim_{h\to 0} \left|\frac{1-h+h}{\lambda-h+0}\right| = \frac{1}{|\lambda|}$. For limit to exist: $|\lambda - 1| = |\lambda| \Rightarrow \lambda = \frac{1}{2}$. Then $L = \frac{1}{\lambda} = 2$.
BETA
