If $$\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sin \,nx} \right)\left[ {\left( {a - n} \right)nx - \tan \,x} \right]}}{{{x^2}}} = 0,$$        then the value of $$a = ?$$

Solution :
$$\eqalign{ & {\text{Let }}\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sin \,nx} \right)\left[ {\left( {a - n} \right)nx - \tan \,x} \right]}}{{{x^2}}} = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\left( {nx - \frac{{{n^3}{x^3}}}{{3!}}} \right)\left[ {n\left( {a - n} \right)x - \left\{ {x + \frac{{{x^3}}}{3} + .....} \right\}} \right]}}{{{x^2}}} = 0 \cr & \left( {{\text{By using expansion of sin }}x{\text{ and tan }}x} \right) \cr & \Rightarrow {n^2}\left( {a - n} \right) - n = 0 \cr & \Rightarrow an - {n^2} - 1 = 0 \cr & \Rightarrow a = \frac{{{n^2} + 1}}{n} \cr & \Rightarrow a = n + \frac{1}{n} \cr} $$