Continuity and Differentiability
Limits and Expansion
IIT-JEE 2009 Paper 1
Grade 12
Question:
Let $L = \lim_{x \to 0} \frac{a - \sqrt{a^2 - x^2} - \frac{x^2}{4}}{x^4}$, $a > 0$. If $L$ is finite, then:
(1) $a = 2, L = \frac{1}{64}$
(2) $a = 2, L = -\frac{1}{64}$
(3) $a = \frac{1}{2}, L = \frac{1}{4}$
(4) $a = \frac{1}{2}, L = -\frac{1}{4}$
Step-by-Step Solution
Key Concept: Expand $\sqrt{a^2 - x^2} \approx a - \frac{x^2}{2a} - \frac{x^4}{8a^3}$. Numerator $\approx \frac{x^2}{2a} - \frac{x^2}{4} - \frac{x^4}{8a^3}$. For $L$ finite, the $x^2$ coefficient must vanish: $\frac{1}{2a} = \frac{1}{4} \Rightarrow a = 2$. Then $L = \frac{-1}{8 \cdot 8} = \frac{1}{64}$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)