If $n0$ and $\lim _{x ightarrow 0} \frac{((a-n) n x- an x) \sin n x}{x^2}=0$, then minimum value of $a$ is

Question

If $n>0$ and $\lim _{x ightarrow 0} \frac{((a-n) n x-	an x) \sin n x}{x^2}=0$, then minimum value of $a$ is

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \lim _{x ightarrow 0} \frac{((a-n) n x-	an x) \sin n x}{x^2}=0 \ & \lim _{x ightarrow 0}\left[\frac{(a-n) n x-	an x}{x}ight] \lim _{x ightarrow 0} \frac{\sin n x}{n x} n=0 \ & \Rightarrow n\left[\lim _{x ightarrow 0} \frac{(a-n) n x}{x}-\lim _{x ightarrow 0} \frac{	an x}{x}ight] \lim _{n x ightarrow 0} \frac{\sin n x}{n x}=0 . \ & \Rightarrow n[(a-n) n-1] 	imes 1=0 \ & \Rightarrow \quad(a-n) n=1 \ & {[\because n 
eq 0]} \ & a=\frac{1}{n}+n \ & a=\frac{n^2+1}{n} \ & \end{aligned} $ $a$ is minimum when $n$ is minimum. $	herefore$ The minimum value of $n$ is 1 $ a=\frac{\mathrm{I}+1}{\mathrm{l}}=2 $

If $n>0$ and $\lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0$, then minimum value of $a$ is

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