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)

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

$L=\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}$
$=\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}}{x^4}-\frac{1}{4x^2}$
$=\underset{x\to 0}{lim}\frac{a^2-(a^2-x^2)}{x^4(a+\sqrt{a^2-x^2})}-\frac{1}{4x^2}$
$=\underset{x\to 0}{lim}\frac{1}{x^2(a+\sqrt{a^2-x^2})}-\frac{1}{4x^2}$
$=\underset{x\to 0}{lim}\frac{4-a-\sqrt{a^2-x^2}}{4x^2(a+\sqrt{a^2-x^2})}$
Now, for L to be finite, numerator must tend to 0.
So, by options if $a=2$, numerator tends to 0.
$\Rightarrow L=\frac{1}{64}$