If $\underset{x→0}{\lim}\frac{((a-n)nx-\tan x)\sin nx}{x^2}=0$, where n is non zero real number, then a is equal to

0 $\frac{n+1}{n}$ $n$ $n+\frac{1}{n}$

$n+\frac{1}{n}$

$\underset{x→0}{\lim}\frac{((a-n)nx-\tan x)\sin nx}{x^2}=0$ $⇒\underset{x→0}{\lim}n((a-n)n-\frac{\tan x}{x})\frac{\sin nx}{nx}=0$ $⇒n((a-n)n-1)=0$ $⇒a-n=\frac{1}{n}$ so $a=n+\frac{1}{n}$