Limits & Continuity
Limit of sec⁻¹ expression; Taylor expansion
Grade Class 12
Question:
If $k = \lim_{x\to 1} \sec^{-1}\!\left(\dfrac{\lambda^2}{\ln x} - \dfrac{\lambda^2}{x-1}\right)$ exists, then the minimum value of $[|\lambda|]$ is .......... (where $[\cdot]$ denotes GIF)
Step-by-Step Solution
Key Concept: Substitute $x=1+t$, $t\to0$. Simplify $\frac{\lambda^2}{\ln(1+t)} - \frac{\lambda^2}{t}$ using $\ln(1+t)\approx t-t^2/2+\ldots$
Limit equals $\sec^{-1}(\lambda^2/2)$. Requires $\lambda^2/2\geq1$, so $|\lambda|\geq\sqrt{2}$, minimum $[|\lambda|]=1$.
Correct Answer: 1