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

Solution:

We have,

$\begin{array}{l}L=\underset{x\to 0}{lim} \frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}\\ \Rightarrow L=\underset{x\to 0}{lim} \frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}\\ \Rightarrow L=\underset{x\to 0}{lim} \frac{1}{x^2}\left\{\frac{1}{a+\sqrt{a^2-x^2}}-\frac{1}{4}\right\}\end{array}$

$\Rightarrow L=\underset{x\to 0}{lim} \frac{4-a-\sqrt{a^2-x^2}}{4x^2\left(a+\sqrt{\left.a^2-x^2\right)}\right.}$

It is given that $L$ is finite.

$\underset{x\to 0}{lim} 4-a-\sqrt{a^2-x^2}-0\Rightarrow 4-a-a=0\Rightarrow a=2$

Putting $a=2$, we obtain

$\begin{array}{l}L=\underset{x\to 0}{lim} \frac{2-\sqrt{4-x^2}}{4x^2\left(2+\sqrt{4-x^2}\right)}\\ \Rightarrow L=\underset{x\to 0}{lim} \frac{4-\left(4-x^2\right)}{4x^2{\left(2+\sqrt{4-x^2}\right)}^2}\\ \Rightarrow L=\underset{x\to 0}{lim} \frac{1}{4{\left(2+\sqrt{4-x^2}\right)}^2}=\frac{1}{64}\end{array}$