Question

Let $L=\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}-x^2/4}{x^4},a>0$.

If $L$ is finite, then?

• A. $a=2,L=\frac{1}{64}$
• B. $a=1,L=\frac{1}{64}$
• C. $a=3,L=\frac{1}{32}$
• D. $a=1,L=\frac{1}{32}$

Answer 1

Answer: (A)

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

$L=\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}-x^2/4}{x^4}$
Applying L-Hospital's rule, $L=\underset{x\to 0}{lim}\frac{2x-x\sqrt{a^2-x^2}}{8x^3\sqrt{a^2-x^2}}$
$=\underset{x\to 0}{lim}\frac{x\sqrt{a^2-x^2}(\frac{2}{\sqrt{a^2-x^2}}-1)}{8x^3\sqrt{a^2-x^2}}$
$=\underset{x\to 0}{lim}\frac{(\frac{2}{\sqrt{a^2-x^2}}-1)}{8x^2}$
As $x\to 0$, denominator tends to $0$, so the numerator also tends to $0$.
$\Rightarrow \underset{x\to 0}{lim}\frac{2}{\sqrt{a^2-x^2}}-1=0$
$\Rightarrow a=2$
So, $L=\underset{x\to 0}{lim}\frac{2-\sqrt{4-x^2}-x^2/4}{x^4}$
Applying L-Hospital's rule, $=\underset{x\to 0}{lim}\frac{(\frac{2}{\sqrt{4-x^2}}-1)}{8x^2}$
Again ,applying L-Hospital's rule, $=\underset{x\to 0}{lim}\frac{2x{(4-x^2)}^{-3/2}}{16x}=\frac{1}{64}$
Hence, option 'A' is correct.