Question

If the value of $\underset{x\to 0}{lim}\frac{\left(\right(a-n)nx-\tan x)\sin nx}{x^2}$ is equal to $0$, where $n\in R-\left\{0\right\}$, then the value of $a$ is equal to

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

Answer 1

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

The given limit can be written as

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

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

$\Rightarrow \left(1\right)\left(n\right)\left(\right(a-n)n-1)=0$.............$({lim}_{x\to 0}\frac{\sin nx}{nx}=0)$

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