Let $k>0$ and $\lambda =\underset{x\to 0}{lim} \frac{k\left(1-4\sqrt{k^2-x^2}\right)}{x^2\sqrt{k^2-x^2}}$ be finite. Then the value of $\lambda k,$ is

Solution:

It is given that

       $\lambda =\underset{x\to 0}{lim} \frac{k\left(1-4\sqrt{k^2-x^2}\right)}{x^2\sqrt{k^2-x^2}}$ exists

$\begin{array}{l}\Rightarrow \underset{x\to 0}{lim} k\left(1-4\sqrt{k^2-x^2}\right)=0\Rightarrow 1-4k=0\Rightarrow k=\frac{1}{4}\\ \therefore \lambda =\underset{x\to 0}{lim} \frac{1-\sqrt{1-16x^2}}{x^2\sqrt{1-16x^2}}\end{array}$

$\begin{array}{l}\Rightarrow \lambda =\underset{x\to 0}{lim} \frac{1-\left(1-16x^2\right)}{x^2\left(1+\sqrt{1-16x^2}\right)\sqrt{1-16x^2}}\\ \Rightarrow \lambda =16\underset{x\to 0}{lim} \frac{1}{\left(1+\sqrt{1-16x^2}\right)\sqrt{1-16x^2}}=8\end{array}$

Hence, $\lambda =8$ and $k=\frac{1}{4}$