Question

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

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

Answer 1

Answer: (A), (C)

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

Applying L-Hospital's rule
$L=\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}=\underset{x\to 0}{lim}\frac{x{(a^2-x^2)}^{-1/2}-\frac{x}{2}}{4x^3}=\underset{x\to 0}{lim}\frac{{(a^2-x^2)}^{-1/2}+x^2{(a^2-x^2)}^{-3/2}-\frac{1}{2}}{12x^2}$
This is finite if $1/a-1/2=0$ (as numerator tends to $1/a-1/2$ as x tends to 0), so $a=2$
Again applying L-Hospital's rule
$\underset{x\to 0}{lim}\frac{x{(a^2-x^2)}^{-3/2}+2x{(a^2-x^2)}^{-3/2}+3x^3{(a^2-x^2)}^{-3/2}}{24x}=\underset{x\to 0}{lim}\frac{3{(a^2-x^2)}^{-3/2}+3x^2{(a^2-x^2)}^{-3/2}}{24}=\frac{1}{8a^3}=\frac{1}{64}$