Prove the following


$f(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & , \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-1} & , \text { if } 0 \leq x \leq 1\end{array}\right.$ at $x=0$


Finding the left hand and right hand limits for the given function, we have

$\lim _{x \rightarrow 0^{-}} f(x)=\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}$

$=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \times \frac{\sqrt{1+k x}+\sqrt{1-k x}}{\sqrt{1+k x}+\sqrt{1-k x}}$

$=\lim _{x \rightarrow 0^{-}} \frac{(1+k x)-(1-k x)}{x[\sqrt{1+k x}+\sqrt{1-k x}]}$

$=\lim _{x \rightarrow 0^{-}} \frac{1+k x-1+k x}{x[\sqrt{1+k x}+\sqrt{1-k x}]}$

$=\lim _{x \rightarrow 0^{\circ}} \frac{2 k x}{x[\sqrt{1+k x}+\sqrt{1-k x}]}$

$=\lim _{x \rightarrow 0^{-}} \frac{2 k}{\sqrt{1+k x}+\sqrt{1-k x}}$

$=\lim _{h \rightarrow 0} \frac{2 k}{\sqrt{1+k(0-h)}+\sqrt{1-k(0-h)}}$

$=\frac{2 k}{\sqrt{1}+\sqrt{1}}=\frac{2 k}{2}=k$

$\lim _{x \rightarrow 0} f(x)=\frac{2 x+1}{x-1}=\frac{2(0)+1}{0-1}=\frac{1}{-1}=-1$

As the function is continuous at $x=0$.

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} f(x)$


Therefore, the value of k is -1

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