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

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(A) Given that $f(x)$ is continuous on $[-1, 1]$, it must be continuous at $x = 0$.Therefore, $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.F

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$-1$

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(A) Given that $f(x)$ is continuous on $[-1, 1]$, it must be continuous at $x = 0$.Therefore, $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First, calculate the Left Hand Limit $(LHL)$ at $x = 0$:$\lim_{x 	o 0^-} \frac{\sqrt{1+ax}-\sqrt{1-ax}}{x} = \lim_{x 	o 0^-} \frac{(\sqrt{1+ax}-\sqrt{1-ax})(\sqrt{1+ax}+\sqrt{1-ax})}{x(\sqrt{1+ax}+\sqrt{1-ax})}$$= \lim_{x 	o 0^-} \frac{(1+ax)-(1-ax)}{x(\sqrt{1+ax}+\sqrt{1-ax})} = \lim_{x 	o 0^-} \frac{2ax}{x(\sqrt{1+ax}+\sqrt{1-ax})} = \frac{2a}{1+1} = a$.Next, calculate the Right Hand Limit $(RHL)$ at $x = 0$:$\lim_{x 	o 0^+} \frac{x^2+2}{x-2} = \frac{0^2+2}{0-2} = \frac{2}{-2} = -1$.Since the function is continuous, $LHL$ = $RHL$, so $a = -1$.

If $f(x) = \begin{cases} \frac{\sqrt{1+ax}-\sqrt{1-ax}}{x}, & -1 \leq x < 0 \\ \frac{x^2+2}{x-2}, & 0 \leq x \leq 1 \end{cases}$ is continuous on $[-1,1]$, then $a=$

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