Question

Find the value of $k$, for which $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}& if-1\le x<0\\ \frac{2x+1}{x-1}& if0\le x<1\end{array}\right\}$ is continuous at $x=0$.

Answer 1

The given function f(x) is continuous at x = 0.

$\underset{x\to 0}{lim}f\left(x\right)$

$\frac{0+1}{0-1}=\underset{x\to 0}{lim}\left(\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\right)$

$\Rightarrow -1=\underset{x\to 0}{lim}\left(\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\right)\left(\frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}}\right)$

$\Rightarrow -1=\underset{x\to 0}{lim}\left(\frac{1+kx-1+kx}{x[\sqrt{1+kx}+\sqrt{1-kx}]}\right)$

$\Rightarrow -1=\underset{x\to 0}{lim}\left(\frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}}\right)$

When $k=-1$

$\Rightarrow -1=\frac{2k}{2}$

$k=-1$

Therefore the value of $k$ is $-1.$