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 $x=0$

Continuity of Functions : Example #1 - find the value of k

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
$k=-1$

3 Comments

If you have any question, please let me know by comment or email. Also, if you want answers of any mathematical problem, please comment me the question. I will post the answer as early as possible.

Post a Comment

If you have any question, please let me know by comment or email. Also, if you want answers of any mathematical problem, please comment me the question. I will post the answer as early as possible.

Previous Post Next Post