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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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The correct answer is:
2
$$
\begin{aligned}
& \text { } \lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0 \\
& \lim _{x \rightarrow 0}\left[\frac{(a-n) n x-\tan x}{x}\right] \lim _{x \rightarrow 0} \frac{\sin n x}{n x} n=0 \\
& \Rightarrow n\left[\lim _{x \rightarrow 0} \frac{(a-n) n x}{x}-\lim _{x \rightarrow 0} \frac{\tan x}{x}\right] \lim _{n x \rightarrow 0} \frac{\sin n x}{n x}=0 . \\
& \Rightarrow n[(a-n) n-1] \times 1=0 \\
& \Rightarrow \quad(a-n) n=1 \\
& {[\because n \neq 0]} \\
& a=\frac{1}{n}+n \\
& a=\frac{n^2+1}{n} \\
&
\end{aligned}
$$
$a$ is minimum when $n$ is minimum.
$\therefore$ The minimum value of $n$ is 1
$$
a=\frac{\mathrm{I}+1}{\mathrm{l}}=2
$$
\begin{aligned}
& \text { } \lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0 \\
& \lim _{x \rightarrow 0}\left[\frac{(a-n) n x-\tan x}{x}\right] \lim _{x \rightarrow 0} \frac{\sin n x}{n x} n=0 \\
& \Rightarrow n\left[\lim _{x \rightarrow 0} \frac{(a-n) n x}{x}-\lim _{x \rightarrow 0} \frac{\tan x}{x}\right] \lim _{n x \rightarrow 0} \frac{\sin n x}{n x}=0 . \\
& \Rightarrow n[(a-n) n-1] \times 1=0 \\
& \Rightarrow \quad(a-n) n=1 \\
& {[\because n \neq 0]} \\
& a=\frac{1}{n}+n \\
& a=\frac{n^2+1}{n} \\
&
\end{aligned}
$$
$a$ is minimum when $n$ is minimum.
$\therefore$ The minimum value of $n$ is 1
$$
a=\frac{\mathrm{I}+1}{\mathrm{l}}=2
$$
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