Let $$ L=\lim _{x \rightarrow 0} \frac{a-\sqrt{a^{2}-x^{2}}-\frac{x^{2}}{4}}{x^{4}}, \quad a>0 . $$ If $L$ is finite, then

(A) $a=2$; (C) $L=\frac{1}{64}$

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Limits XII

Let

L = x \to 0 lim \frac{a - a ^{2} - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}, a > 0.

L

is finite, then

Answer: (A), (C)

Rationalizing the numerator:

L = x \to 0 lim \frac{a ^{2} - ( a ^{2} - x ^{2} )}{x ^{4} ( a + a ^{2} - x ^{2} )} - \frac{1}{4 x ^{2}}

= x \to 0 lim \frac{x ^{2}}{x ^{4} ( a + a ^{2} - x ^{2} )} - \frac{1}{4 x ^{2}}

= x \to 0 lim \frac{1}{x ^{2}} [\frac{1}{a + a ^{2} - x ^{2}} - \frac{1}{4}]

For the limit to be finite, the expression in brackets must approach zero as

x \to 0

\frac{1}{2 a} = \frac{1}{4} \Rightarrow a = 2

With

a = 2

, applying L'Hôpital's rule or Taylor expansion:

L = x \to 0 lim \frac{2 - 4 - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}

Using Taylor expansion

4 - x^{2} = 2 - \frac{x ^{2}}{4} - \frac{x ^{4}}{64} + O (x^{6})

L = x \to 0 lim \frac{\frac{x ^{4}}{64}}{x ^{4}} = \frac{1}{64}

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · One or more correct options