$f(x)=\left\{\begin{array}{cc}
\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} & -1 \leq x<0 \\
\frac{2 x+1}{x-2} & 0 \leq x \leq 1
\end{array}\right.$

is continuous in the interval $[-1,1]$, then $p$ is equal to :

$-\frac{1}{2}$

$f(x)=\left\{\begin{array}{cl}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} & -1 \leq x<0 \\ \frac{2 x+1}{x-2} & 0 \leq x \leq 1\end{array}\right.$

f is continous

⇒ $\lim\limits_{x→0^-}f((x) = \lim\limits_{x→0^+}f(x) = f(0)$

computing right hand limit

RHD = $\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0} \frac{2 x+1}{x-2}=\frac{0+1}{0-2}=\frac{-1}{2}$

$f(0)=-1 / 2$

computing left hand limit

LHD = $\lim\limits_{x \rightarrow 0^{-}}(f-x)=\lim\limits_{x \rightarrow 0} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}$

$=\lim\limits_{x \rightarrow 0} \frac{\left(\sqrt{1+p x}-\sqrt{1-p^x}\right)\left(\sqrt{1+p x}+\sqrt{1-p^x}\right)}{x\left(\sqrt{1+p^x}+\sqrt{1-p x}\right)}$

$=\lim\limits_{n \rightarrow 0} \frac{1+p x-1+p x}{x(\sqrt{1+p x}+\sqrt{1-p x})}$

$=\lim\limits_{x \rightarrow 0} \frac{2 p x}{x(\sqrt{1+p x}+\sqrt{1-p x})}$

LHD = $\lim\limits_{x \rightarrow 0} \frac{2 p}{\sqrt{1+p x}+\sqrt{1-p x}}=\frac{2 p}{\sqrt{1+0}+\sqrt{1-0}}$

LHD = $\frac{2}{2}=p$

LHD = RHS = f(0) ⇒  p = $\frac{-1}{2}$