The values of $p$and $q$for which the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sin (p+1)x+\sin x}{x}& x<0\\ q,& x=0\\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3/2}}& x>0\end{array}\right.$

 is continuous for all $x$are

 $$\begin{gathered} At x=0,f\left(0\right)=q \\ \text{For }x<0, \\ L.H.L=\underset{x\to 0}{lim} f\left(x\right)=\underset{x\to 0}{lim}\frac{\sin (p+1)x+\sin x}{x} \\ =\underset{x\to 0}{lim}\frac{\sin (p+1)x}{x}+\underset{x\to 0}{lim}\frac{\sin x}{x} \\ =p+1+1 \\ =p+2 \\ for x>0 \\ R.H.L=\underset{x\to 0}{\lim }\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{{\displaystyle \frac{3}{2}}}} \\ =\underset{x\to 0}{\lim }\frac{\sqrt{x}(\sqrt{1+x}-1)}{x\sqrt{x}} \\ =\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1}{x}\times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} \\ =\underset{x\to 0}{\lim }\frac{x}{x(\sqrt{1+x}+1)} \\ =\frac{1}{2} \\ Since f\left(x\right) is continuous at x=0 then \\ L.H.L=R.H.L=f\left(0\right) \\ \Rightarrow p+2=\frac{1}{2}=q \\ \Rightarrow p=\frac{-3}{2},q=\frac{1}{2} \end{gathered}$$

$\because f\left(x\right)\text{ is continuous }\Rightarrow p=-3/2,q=1/2$