If undersetx→0lim(((a-n)nx-tan x)sin nx)/(x^2)=0, where n is non zero…

$\underset{x→0}{\lim}\frac{((a-n)nx-\tan x)\sin nx}{x^2}=0$

$⇒\underset{x→0}{\lim}n((a-n)n-\frac{\tan x}{x})\frac{\sin nx}{nx}=0$

$⇒n((a-n)n-1)=0$

$⇒a-n=\frac{1}{n}$

so $a=n+\frac{1}{n}$