If f(x) = \ cases ( ( p + 1 ) x + x)/(x) & , & x 0 cases ) is...

∵ f\left(x\right) = \left\{$$\begin{array}{ll}\sin \left(p + 1\right)x + \sin x x\end{array}$$ q \dfrac{\sqrt{{x}^{2} + x} - \sqrt{x}}{{x}^{3 / 2}} $$\begin{array}{ll}: x 0\end{array}$$\right)

$∵ f$ is continuous at $x = 0$

$L . H . L =$

$\lim_{h \to 0}\dfrac{\sin \left(p + 1\right)\left(- h\right) + \sin ( - h )}{- h} = \lim_{h \to 0}\dfrac{\sin \left(p + 1\right)h + sinh}{h}$

$= \lim_{h \to 0}\dfrac{\sin \left(p + 1\right)h}{h\left(p + 1\right)}\left(p + 1\right) + \lim_{h \to 0}\dfrac{sinh}{h} = p + 1 + 1$

$= p + 2$

$R . H . L = \lim_{h \to 0}\dfrac{\sqrt{{h}^{2} + h}-\sqrt{h}}{h\dfrac{3}{2}} \times \dfrac{\sqrt{{h}^{2} + h} + \sqrt{h}}{\sqrt{{h}^{2} + h} + \sqrt{h}} = \lim_{h \to 0}\dfrac{{h}^{2} + h - h}{h\dfrac{3}{2}h\dfrac{1}{2}\left(\sqrt{h + 1} + 1\right)}$

$\lim_{h \to 0}\dfrac{1}{\sqrt{1 + h} + 1} = \dfrac{1}{2}$

$⇒ L . H . L . = R . H . L . = f\left(0\right) ⇒ p + 2 = \dfrac{1}{2} = q$
$=$$\begin{array}{ll}p = - \dfrac{3}{2} \text{ and } q = \dfrac{1}{2}\end{array}$$$