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

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(B) For the limit to exist at $x=0$,the Left Hand Limit $(LHL)$ must equal the Right Hand Limit $(RHL)$.$RHL = \lim_{x \rightarrow 0^+} \frac{2x+1}{x-

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$-\frac{1}{2}$

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(B) For the limit to exist at $x=0$,the Left Hand Limit $(LHL)$ must equal the Right Hand Limit $(RHL)$.$RHL = \lim_{x \rightarrow 0^+} \frac{2x+1}{x-2} = \frac{2(0)+1}{0-2} = -\frac{1}{2}$.$LHL = \lim_{x \rightarrow 0^-} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}$.Using rationalization:$LHL = \lim_{x \rightarrow 0^-} \frac{(\sqrt{1+px} - \sqrt{1-px})(\sqrt{1+px} + \sqrt{1-px})}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{(1+px) - (1-px)}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{2px}{x(\sqrt{1+px} + \sqrt{1-px})} = \frac{2p}{1+1} = p$.Equating $LHL = RHL$,we get $p = -\frac{1}{2}$.

Define $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$. If $\lim_{x \rightarrow 0} f(x)$ exists,then $p =$

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