Question

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

We have
$\underset{x\to 0}{lim}\{\frac{(a-n)nx}{x}-\frac{\tan x}{x}\}\frac{\sin nx}{x}=0$

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

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

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