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

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

Vedclass · Accepted Answer

$-2$

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(B) For $f(x)$ to be continuous at $x = 0$, the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$ and the value of the function $f(0)$.$LHL$ = $\lim_{x 	o 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$Rationalizing the numerator:$LHL$ = $\lim_{x 	o 0^-} \frac{(\sqrt{1+kx}-\sqrt{1-kx})(\sqrt{1+kx}+\sqrt{1-kx})}{x(\sqrt{1+kx}+\sqrt{1-kx})}$$LHL$ = $\lim_{x 	o 0^-} \frac{(1+kx)-(1-kx)}{x(\sqrt{1+kx}+\sqrt{1-kx})} = \lim_{x 	o 0^-} \frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})} = \frac{2k}{1+1} = k$$RHL$ = $\lim_{x 	o 0^+} (2x^2+3x-2) = 2(0)^2+3(0)-2 = -2$Since the function is continuous, $LHL$ = $RHL$, therefore $k = -2$.

If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & -1 \leq x < 0 \\ 2x^2+3x-2, & 0 \leq x \leq 1 \end{cases}$ is continuous at $x = 0$, then $k$ is equal to

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