# Find the value of $k$, for which $f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, \text{if} \ -1\le x < 0 \\ \frac{2x+1}{x-1}, \text{if} \ 0\le x < 1 \end{cases}$ is continuous at $x=0$

## Example:

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

### Solution:

Given that,
$f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, \text{if} \ -1\le x<0 \\ \frac{2x+1}{x-1}, \text{if} \ 0\le x<1 \end{cases}$
Since, f(x) is continuous at $x=0$,
$\displaystyle\lim_{x \to 0^-} f(x)=f(0)=\displaystyle\lim_{x \to 0^+} f(x)$ ............ (1)
Here, $f(0)=\frac{2(0)+1}{0-1}$
$=\frac{1}{-1}$
$f(0)=-1$              ..................... (2)
Now, $\displaystyle\lim_{x \to 0^-} f(x)=\displaystyle\lim_{x \to 0}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$
$=\displaystyle\lim_{x \to 0}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\times \frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}}$
$=\displaystyle\lim_{x \to 0}\frac{\{1+kx\}-\{1-kx\}}{x(\sqrt{1+kx}+\sqrt{1-kx})}$
$=\displaystyle\lim_{x \to 0}\frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})}$
$=\displaystyle\lim_{x \to 0}\frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}}$
$=\frac{2k}{\sqrt{1+0}+\sqrt{1-0}}$
$=\frac{2k}{2}$
$\therefore \displaystyle\lim_{x \to 0^-} f(x)=k$      ..................... (3)
From (1), (2) and (3), we have
$k=-1$