If f(x) = begin{cases} frac{sqrt{1+kx}-sqrt{1 | vedclass

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(B) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$ at $x=0$. First,calculate th

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$k=-1$

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(B) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$ at $x=0$. First,calculate the $LHL$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$ Multiply the numerator and denominator by the conjugate $(\sqrt{1+kx} + \sqrt{1-kx})$: $= \lim_{x 	o 0^-} \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x 	o 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x 	o 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} = \frac{2k}{1+1} = k$. Next,calculate the $RHL$: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{2x+1}{x-1} = \frac{2(0)+1}{0-1} = -1$. Since the function is continuous at $x=0$,we set $LHL$ = $RHL$: $k = -1$.

If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x} & ; -1 \leq x < 0 \\ \frac{2x+1}{x-1} & ; 0 \leq x \leq 1 \end{cases}$ is continuous at $x=0$,then the value of $k$ is:

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