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), (C)

The correct options are
A $a=2$
C $L=\frac{1}{64}$

$L={lim}_{x\to 0}\frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}$

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

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

$={lim}_{x\to 0}\frac{a-a(1-\frac{1}{2}\frac{x^2}{a^2}-\frac{1}{8}\frac{x^4}{a^4})-\frac{x^2}{4}}{x^4}$

$={lim}_{x\to 0}\frac{\frac{1}{2}\frac{x^2}{a}+\frac{1}{8}\frac{x^4}{a^2}-\frac{x^2}{4}}{x^4}$

$={lim}_{x\to 0}\frac{x^2(\frac{1}{2a}-\frac{1}{4})+\frac{x^4}{8a^3}}{x^4}$

Limit is finite, when $\frac{1}{2a}=\frac{1}{4}$

$a=2$

$\therefore L=\frac{1}{8a^3}$

$\therefore L=\frac{1}{8x^4}$

$\therefore L=\frac{1}{64}$

$\therefore L=\frac{1}{64},a=2$