JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 17)
Let $$L = \mathop {\lim }\limits_{x \to 0} {{a - \sqrt {{a^2} - {x^2}} - {{{x^2}} \over 4}} \over {{x^4}}},a > 0$$. If L is finite, then
$$a = 2$$
$$a = 1$$
$$L = {1 \over {64}}$$
$$L = {1 \over {32}}$$
Explanation
The given limit is
$$L = \mathop {\lim }\limits_{x \to 0} {{a - \sqrt {{a^2} - {x^2}} - {{{x^2}} \over 4}} \over {{x^4}}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}(a + \sqrt {{a^2} - {x^2}} )}} - {1 \over {4{x^2}}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{(4 - a) - \sqrt {{a^2} - {x^2}} } \over {4{x^2}(a + \sqrt {{a^2} - {x^2}} )}}$$
The numerator $$\to$$ 0 if $$a = 2$$; therefore, $$L = {1 \over {64}}$$.
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