Let \[f\left( x \right) = \left\{ \begin{array}{l} \frac{{1 - {{\sin }^3}x}}{{3\,{{\cos }^2}x}},\,\,\,\,\,\,\,x < \frac{\pi }{2}\\ \,\,\,\,\,\,\,p,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{\pi }{2}\\ \frac{{q\left( {1 - \sin \,x} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}},\,\,x > \frac{\pi }{2} \end{array} \right.\]
If $$f\left( x \right)$$  is continuous at $$x = \frac{\pi }{2},\,\left( {p,\,q} \right) = ?$$

Solution :
$$\eqalign{ & f\left[ {{{\left( {\frac{\pi }{2}} \right)}^ - }} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{1 - {{\sin }^3}\left[ {\left( {\frac{\pi }{2}} \right) - h} \right]}}{{3\,{{\cos }^2}\left[ {\left( {\frac{\pi }{2}} \right) - h} \right]}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - {{\cos }^3}h}}{{3\,{{\sin }^2}h}} \cr & = \frac{1}{2} \cr & f\left[ {{{\left( {\frac{\pi }{2}} \right)}^ + }} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{q\left[ {1 - \sin \left\{ {\left( {\frac{\pi }{2}} \right) + h} \right\}} \right]}}{{{{\left[ {\pi - 2\left\{ {\left( {\frac{\pi }{2}} \right) + h} \right\}} \right]}^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{q\left( {1 - \cos \,h} \right)}}{{4\,{h^2}}} \cr & = \frac{q}{8} \cr & \therefore \,p = \frac{1}{2} = \frac{q}{8} \cr & \Rightarrow p = \frac{1}{2},\,\,q = 4 \cr} $$