$$f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {x < 0} \cr q & {x = 0} \cr {{{\sqrt {{x^2} + x} - \sqrt x } \over {{x^{{3 \over 2}}}}}} & {x > 0} \cr } } \right.$$

is continuous at x = 0

So f(0^–) = f(0) = f (0⁺) ... (1)

$$f({0^ - }) = \mathop {lt}\limits_{h \to 0} f(0 - h)$$

$$ \Rightarrow \mathop {lt}\limits_{h \to 0} {{\sin (p + 1)( - h) + \sin ( - h)} \over { - h}}$$

$$ \Rightarrow \mathop {lt}\limits_{h \to 0} \left[ {{{ - \sin (p + 1)h} \over { - h}} + {{\sinh } \over h}} \right]$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sin (p + 1)h} \over {h(p + 1)}} \times (p + 1) + \mathop {\lim }\limits_{h \to 0} {{\sinh } \over h}$$

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

Now $$f({0^ + }) = \mathop {\lim }\limits_{h \to 0} (0 + h) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {{h^2} + h} - \sqrt h } \over {{h^{3/2}}}}$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(h)}^{{1 \over 2}}}\left[ {\sqrt {h + 1} - 1} \right]} \over {h\left( {{h^{{1 \over 2}}}} \right)}}$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sqrt {h + 1} - 1} \over h} \times {{\sqrt {h + 1} + 1} \over {\sqrt {h + 1} + 1}}$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{h + 1 - 1} \over {h\left( {\sqrt {h + 1} + 1} \right)}}$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {1 \over {\sqrt {h + 1} + 1}} = {1 \over {1 + 1}} = {1 \over 2}$$ ..... (3)

Now, from equation (1)
f(0^–) = f(0) = f(0⁺)

p + 2 = q = 1/2

So, $$q = {1 \over 2}$$ and $$p = {1 \over 2} - 2 = {{ - 3} \over 2}$$

$$(p,q) \equiv \left( { - {3 \over 2},{1 \over 2}} \right)$$