is continuous at $x=0$, then the ordered pair $(p, q)$ is equal to:

$f(x)=\left\{\begin{array}{ll}\frac{\sin (\mathrm{p}+1) \mathrm{x}+\sin \mathrm{x}}{\mathrm{x}} & \mathrm{x}<0 \\ \mathrm{q} & \mathrm{x}=0 \\ \frac{\sqrt{\mathrm{x}^{2}+\mathrm{x}}-\sqrt{\mathrm{x}}}{\mathrm{x}^{\frac{3}{2}}} & \mathrm{x}>0\end{array}\right.$ is continuous at $x=0$

Therefore, $f\left(0^{-}\right)=f(0)=f\left(0^{+}\right)$.......(1)

$f\left(0^{-}\right)=\operatorname{Lim}_{h \rightarrow 0} f(0-h)=\operatorname{Lim}_{h \rightarrow 0} \frac{\sin (p+1)(-h)+\sin (-h)}{-h}$

$=\operatorname{Lim}_{h \rightarrow 0}\left[\frac{-\sin (p+1) h}{-h}+\frac{\sin h}{h}\right]$

$=\operatorname{Lim}_{h \rightarrow 0} \frac{\sin (p+1) h}{h(p+1)} \times(p+1)+\operatorname{Lim}_{h \rightarrow 0} \frac{\sin h}{h}$

$=(p+1)+1=p+2$......(2)

And $f\left(0^{+}\right)=\operatorname{Lim}_{\mathrm{h} \rightarrow 0} f(0+h)=\frac{\sqrt{h^{2}+h}-\sqrt{h}}{h^{3 / 2}}$

$=\operatorname{Lim}_{h \rightarrow 0} \frac{(h)^{\frac{1}{2}}[\sqrt{h+1}-1]}{h\left(h^{\frac{1}{2}}\right)}$

$=\operatorname{Lim}_{h \rightarrow 0} \frac{\sqrt{h+1}-1}{h} \times \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}=\operatorname{Lim}_{h \rightarrow 0} \frac{h+1-1}{h(\sqrt{h+1}+1)}$

$=\operatorname{Lim}_{h \rightarrow 0} \frac{1}{\sqrt{h+1}+1}=\frac{1}{1+1}=\frac{1}{2}$.....(3)

Now, from equation (1),

$f\left(0^{-}\right)=f(0)=f\left(0^{+}\right) \Rightarrow p+2=q=\frac{1}{2}$

$\Rightarrow q=\frac{1}{2}$ and $p=\frac{-3}{2} \quad \therefore(p, q)\left(-\frac{3}{2}, \frac{1}{2}\right)$