Let $K0$ and $\lambda=\lim _{x \rightarrow 0} \frac{K\left(1-4 \sqrt{K^2-x^2}\right)}{x^2 \sqrt{K^2

Question

Let $K>0$ and $\lambda=\lim _{x \rightarrow 0} \frac{K\left(1-4 \sqrt{K^2-x^2}\right)}{x^2 \sqrt{K^2-x^2}}$ is finite then the value of $\lambda \mathrm{K}$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

In given limit as. $x ightarrow 0 \Rightarrow D^r ightarrow 0$ $\  \begin{aligned} & \Rightarrow N^r ightarrow 0 \ & \Rightarrow \quad k(1-4 k)=0 \ & \Rightarrow k=\frac{1}{4} \quad \because k>0 \ & \Rightarrow \quad \lambda=\lim _{x ightarrow 0}\left(\frac{1-\sqrt{1-16 x^2}}{x^2 \sqrt{1-16 x^2}}ight) \ & \lambda=\lim _{x ightarrow 0^{-}} \frac{1-\left(1-8 x^2-32 x^4ight)}{x^2\left(1-8 x^2ight)} \ & \lambda=\lim _{x ightarrow 0} \frac{x^2\left(8-32 x^2ight)}{x^2\left(1-8 x^2ight)} \Rightarrow \lambda=8 \end{aligned} $