Question

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