If n 0 and lim _{x rightarrow 0} frac{((a-n | vedclass

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(B) Given $\lim _{x ightarrow 0} \frac{((a-n) n x-	an x) \sin n x}{x^2}=0$. We can rewrite the limit as: $\lim _{x ightarrow 0} \left[ \frac{(a-n

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$2$

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(B) Given $\lim _{x ightarrow 0} \frac{((a-n) n x-	an x) \sin n x}{x^2}=0$. We can rewrite the limit as: $\lim _{x ightarrow 0} \left[ \frac{(a-n) n x-	an x}{x} ight] \cdot \lim _{x ightarrow 0} \frac{\sin n x}{x} = 0$. Since $\lim _{x ightarrow 0} \frac{\sin n x}{x} = n$,we have: $n \cdot \lim _{x ightarrow 0} \left[ (a-n) n - \frac{	an x}{x} ight] = 0$. $n [ (a-n) n - 1 ] = 0$. Since $n > 0$,we must have $(a-n) n - 1 = 0$. $(a-n) n = 1 \implies a-n = \frac{1}{n} \implies a = n + \frac{1}{n}$. By $AM-GM$ inequality,for $n > 0$,$n + \frac{1}{n} \ge 2$. The minimum value of $a$ is $2$ when $n = 1$.

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

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