If lim _{x rightarrow 0} frac{[(a-n) n x-tan | vedclass

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(C) Given $\lim _{x ightarrow 0} \frac{((a-n) n x-	an x) \sin n x}{x^2}=0, n 
eq 0$. Dividing the numerator and denominator by $x^2$,we get: $\lim

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$2$

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(C) Given $\lim _{x ightarrow 0} \frac{((a-n) n x-	an x) \sin n x}{x^2}=0, n 
eq 0$. Dividing the numerator and denominator by $x^2$,we get: $\lim _{x ightarrow 0} \left( \frac{(a-n) n x - 	an x}{x} ight) \left( \frac{\sin n x}{x} ight) = 0$. $\lim _{x ightarrow 0} \left( (a-n)n - \frac{	an x}{x} ight) \left( n \cdot \frac{\sin n x}{n x} ight) = 0$. Since $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{\sin n x}{n x} = 1$,we have: $((a-n)n - 1) \cdot n = 0$. Since $n 
eq 0$,we have $(a-n)n - 1 = 0$,which implies $an - n^2 = 1$,or $a = n + \frac{1}{n}$. By the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,for $n > 0$: $\frac{n + \frac{1}{n}}{2} \geq \sqrt{n \cdot \frac{1}{n}} = 1$. Therefore,$a \geq 2$. The minimum possible positive value of $a$ is $2$.

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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