Question

If $\underset{x\to 0}{lim}\frac{\{(a-n)nx-\tan x\}\sin nx}{x^2}=0$, where $n$ is nonzero 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 $n+\frac{1}{n}$

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

$\Rightarrow \{(a-n)n-1\}n=0$
$\Rightarrow (a-n)n=1\Rightarrow a=n+\frac{1}{n}$