If f:[-2,2] rightarrow R is defined by f(x) = | vedclass

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(B) Given that $f:[-2,2] \rightarrow R$ is continuous on $[-2,2]$,it must be continuous at $x=0$. For continuity at $x=0$,we must have $\lim_{x \right

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$3$

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(B) Given that $f:[-2,2] \rightarrow R$ is continuous on $[-2,2]$,it must be continuous at $x=0$. For continuity at $x=0$,we must have $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = f(0)$. First,calculate the Left Hand Limit $(LHL)$: $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \frac{\sqrt{1+cx}-\sqrt{1-cx}}{x}$. Multiplying by the conjugate $\frac{\sqrt{1+cx}+\sqrt{1-cx}}{\sqrt{1+cx}+\sqrt{1-cx}}$,we get: $\lim_{x \rightarrow 0^{-}} \frac{(1+cx)-(1-cx)}{x(\sqrt{1+cx}+\sqrt{1-cx})} = \lim_{x \rightarrow 0^{-}} \frac{2cx}{x(\sqrt{1+cx}+\sqrt{1-cx})} = \frac{2c}{\sqrt{1}+\sqrt{1}} = \frac{2c}{2} = c$. Next,calculate the Right Hand Limit $(RHL)$: $\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} \frac{x+3}{x+1} = \frac{0+3}{0+1} = 3$. Since $f$ is continuous at $x=0$,$LHL = RHL$. Therefore,$c = 3$.

If $f:[-2,2] \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\sqrt{1+cx}-\sqrt{1-cx}}{x} & \text{for } -2 \leq x < 0 \\ \frac{x+3}{x+1} & \text{for } 0 \leq x \leq 2 \end{cases}$ and is continuous on $[-2,2]$,then $c$ is equal to

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