If $\lim _{x \rightarrow 0} \frac{[(a-n) n x-\tan x] \sin n x}{x^2}=0,(n \neq 0)$ then the minimum possible positive value of $a$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0, n \neq 0 \ & \Rightarrow \quad \lim _{x \rightarrow 0}\left(\frac{(a-n) n x}{x}-\frac{\tan x}{x}\right) \frac{\sin (n x)}{x}=0 \ & \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin n x}{n x}=n\right] \ & \Rightarrow \quad n\left(a n-n^2-1\right)=0 \Rightarrow \quad a=n+\frac{1}{n}, n \neq 0 \ & \because \quad \frac{n+\frac{1}{n}}{2} \geq \sqrt{n \frac{1}{n}} \text { (by AM } \geq \text { GM) } \ & \Rightarrow \quad \frac{a}{2} \geq 1 \ & \Rightarrow \quad a \geq 2 \end{aligned}$ $\therefore$ Minimum possible positive value of $a$ is 2. Hence, option (c) is correct.

If \(\lim _{x \rightarrow 0} \frac{[(a-n) n x-\tan x] \sin n x}{x^2}=0,(n \neq 0)\) then the minimum…

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