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

• A. $a = 2$
• B. $a = 1$
• C. $L = \frac{1}{64}$
• D. $L = \frac{1}{32}$

Answer 1

Answer: (A)

L = $lim _{ x \to 0} \frac{ a - \sqrt {a^2 - x^2} - \frac{x^2}{4}}{ x^4} , a > 0$
= $lim _{ x \to 0} \frac{ a - a \Bigg [ 1 - \frac{1}{2} . \frac{x^2}{a^2} + \frac{ \frac{1}{2} \bigg(\frac{1}{2} - 1\bigg)}{2} . \frac{x^4}{a^4} - ..... \Bigg ] - \frac{x^2}{ 4}}{x^4}$
= $lim _{ x \to 0} \frac{ \frac{x^2}{2a} + \frac{1}{8} . \frac{x^4}{a^3}+ .... - \frac{x^2}{4} }{x^4}$
Since, L is finite $ \Rightarrow 2a = 4 \Rightarrow \Rightarrow a = 2 $
$$ L = $ lim_{ x \to 0 } \frac{1}{ 8 . a^3} = \frac{1}{64} $ .