Question

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

• A. $0$
• B. $\frac{n+1}{n}$
• C. $n$
• D. $n+\frac{1}{n}$

Answer 1

The correct option is D $n+\frac{1}{n}$

Given : $\underset{x\to 0}{lim}\frac{\left[\right(a-n)nx-\tan x]\sin nx}{x^2}=0$

$\Rightarrow \underset{x\to 0}{lim}n\frac{\sin nx}{nx}\underset{x\to 0}{lim}\left\{\right(a-n)n-\frac{\tan x}{x}\}=0$

$\Rightarrow n\left\{\right(a-n)n-1\}=0$

$\Rightarrow (a-n)n=1$

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