Let $$f\left( x \right) = \frac{{{{\left( {{e^x} - 1} \right)}^2}}}{{\sin \left( {\frac{x}{a}} \right)\log \left( {1 + \frac{x}{4}} \right)}}$$      for $$x \ne 0,$$  and $$f\left( 0 \right) = 12.$$   If $$f$$ is continuous at $$x = 0,$$  then the value of $$a$$ is equal to :

Solution :
$$\eqalign{ & \mathop {{\text{Lt}}}\limits_{x \to 0} \frac{{{{\left( {{e^x} - 1} \right)}^2}}}{{\sin \left( {\frac{x}{a}} \right)\log \left( {1 + \frac{x}{4}} \right)}} = \mathop {{\text{Lt}}}\limits_{x \to 0} \frac{{\frac{{{{\left( {{e^x} - 1} \right)}^2}}}{x}.{x^2}}}{{\frac{x}{a}.\frac{{\sin \left( {\frac{x}{a}} \right)}}{{\left( {\frac{x}{a}} \right)}}.\frac{{\log \left( {1 + \frac{x}{4}} \right)}}{{\left( {\frac{x}{4}} \right)}}.\frac{x}{4}}} \cr & \Rightarrow 4a = 12 \cr & \Rightarrow a = 3 \cr} $$