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HomeMathematicsCalculusFind the value of $k$, for which $f(x)=begin{cases} frac{sqrt{1+kx}-sqrt{1-kx}}{x}, text{if} -1le...

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

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

Example:

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

Solution:

Given that,
$f(x)=begin{cases} frac{sqrt{1+kx}-sqrt{1-kx}}{x}, text{if} -1le x<0 \ frac{2x+1}{x-1}, text{if} 0le x<1 end{cases}$ 
Since, f(x) is continuous at $x=0$,
$displaystylelim_{x to 0^-} f(x)=f(0)=displaystylelim_{x to 0^+} f(x)$ ………… (1)
Here, $f(0)=frac{2(0)+1}{0-1}$ 
                      $=frac{1}{-1}$ 
           $f(0)=-1$              ………………… (2)

Now, $displaystylelim_{x to 0^-} f(x)=displaystylelim_{x to 0}frac{sqrt{1+kx}-sqrt{1-kx}}{x}$ 
                              $=displaystylelim_{x to 0}frac{sqrt{1+kx}-sqrt{1-kx}}{x}times frac{sqrt{1+kx}+sqrt{1-kx}}{sqrt{1+kx}+sqrt{1-kx}}$ 
                               $=displaystylelim_{x to 0}frac{{1+kx}-{1-kx}}{x(sqrt{1+kx}+sqrt{1-kx})}$ 

                               $=displaystylelim_{x to 0}frac{2kx}{x(sqrt{1+kx}+sqrt{1-kx})}$ 
                               $=displaystylelim_{x to 0}frac{2k}{sqrt{1+kx}+sqrt{1-kx}}$ 

                               $=frac{2k}{sqrt{1+0}+sqrt{1-0}}$ 
                               $=frac{2k}{2}$ 
$therefore displaystylelim_{x to 0^-} f(x)=k$      ………………… (3)
From (1), (2) and (3), we have
$k=-1$

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