If the function f(x) = begin{cases} frac{sqrt | vedclass

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(C) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit,i.e.,$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarr

Vedclass · Accepted Answer

$\frac{-1}{2}$

Vedclass · Answer

(C) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit,i.e.,$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x)$.First,calculate the left-hand limit:$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \frac{\sqrt{1+px}-\sqrt{1-px}}{x}$Multiply the numerator and denominator by the conjugate $\sqrt{1+px} + \sqrt{1-px}$:$= \lim_{x \rightarrow 0^-} \frac{(\sqrt{1+px}-\sqrt{1-px})(\sqrt{1+px}+\sqrt{1-px})}{x(\sqrt{1+px}+\sqrt{1-px})}$$= \lim_{x \rightarrow 0^-} \frac{(1+px)-(1-px)}{x(\sqrt{1+px}+\sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{2px}{x(\sqrt{1+px}+\sqrt{1-px})}$$= \lim_{x \rightarrow 0^-} \frac{2p}{\sqrt{1+px}+\sqrt{1-px}} = \frac{2p}{1+1} = p$.Next,calculate the right-hand limit:$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{2x+1}{x-2} = \frac{2(0)+1}{0-2} = \frac{-1}{2}$.Equating the two limits,we get $p = \frac{-1}{2}$.

If the function $f(x) = \begin{cases} \frac{\sqrt{1+px}-\sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $p = $

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