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

Question

If $\lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0$, where $n$ is non-zero real number, then a is equal to, 0, $\frac{\mathrm{n}+1}{\mathrm{n}}$, $\mathrm{n}$, $\mathrm{n}+\frac{1}{\mathrm{n}}$

MARKS App · Accepted Answer

$\begin{aligned} & \lim _{\mathrm{x} \rightarrow 0} \mathrm{n} \frac{\sin \mathrm{nx}}{\mathrm{nx}} \cdot \lim _{\mathrm{x} \rightarrow 0}\left((\mathrm{a}-\mathrm{n}) \mathrm{n}-\frac{\tan \mathrm{x}}{\mathrm{x}}\right)=0 \ & \Rightarrow \mathrm{n}((\mathrm{a}-\mathrm{n}) \mathrm{n}-1)=0 \ & \Rightarrow(\mathrm{a}-\mathrm{n}) \mathrm{n}=1 \Rightarrow \mathrm{a}=\mathrm{n}+\frac{1}{\mathrm{n}} . \end{aligned}$

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