The value of $$p$$ for which the function \[f\left( x \right) = \left\{ \begin{array}{l} \frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \frac{x}{p}\log \left[ {1 + \frac{{{x^2}}}{3}} \right]}},\,\,\,x \ne 0\\ \,\,\,\,12{\left( {\log \,4} \right)^3},\,\,\,\,\,\,x = 0 \end{array} \right.\]        may be continuous at $$x = 0,$$  is :

Solution :
For $$f\left( x \right)$$  to be continuous at $$x = 0,$$  we should have $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 12{\left( {\log \,4} \right)^3}$$
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{4^x} - 1}}{x}} \right)^3} \times \frac{{\left( {\frac{x}{p}} \right)}}{{\left( {\sin \frac{x}{p}} \right)}} \cdot \frac{{p{x^2}}}{{\log \left( {1 + \frac{1}{3}{x^2}} \right)}} \cr & = {\left( {\log \,4} \right)^3} \cdot 1 \cdot p.\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}}}{{\frac{1}{3}{x^2} - \frac{1}{{18}}{x^4} + .....}}} \right) \cr & = 3p{\left( {\log \,4} \right)^3} \cr & {\text{Hence, }}p = 4 \cr} $$