Question

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

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

Answer 1

The correct option is C $L=\frac{1}{64}$

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

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

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

Since $L$ is finite$\Rightarrow 2a=4$

$\Rightarrow a=2$

$\therefore L={lim}_{x\to 0}\frac{1}{8a^3}={lim}_{x\to 0}\frac{1}{8\times 2^3}=\frac{1}{64}$