If mathop {lim }limits_{x to 0} frac{{[(a - n | vedclass

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(D) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{[(a - n)nx - 	an x]\sin nx}{x^2} = 0$We can rewrite the expression as: $\mathop {\lim }\

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$n + \frac{1}{n}$

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(D) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{[(a - n)nx - 	an x]\sin nx}{x^2} = 0$We can rewrite the expression as: $\mathop {\lim }\limits_{x 	o 0} \left( \frac{(a - n)nx - 	an x}{x} \cdot \frac{\sin nx}{x} ight) = 0$Multiplying and dividing by $n$ in the second term: $\mathop {\lim }\limits_{x 	o 0} \left( ((a - n)n - \frac{	an x}{x}) \cdot n \cdot \frac{\sin nx}{nx} ight) = 0$Since $\mathop {\lim }\limits_{x 	o 0} \frac{	an x}{x} = 1$ and $\mathop {\lim }\limits_{x 	o 0} \frac{\sin nx}{nx} = 1$,the equation becomes:$n((a - n)n - 1) = 0$Since $n 
eq 0$,we have $(a - n)n - 1 = 0$$(a - n)n = 1$$a - n = \frac{1}{n}$$a = n + \frac{1}{n}$

If $\mathop {\lim }\limits_{x \to 0} \frac{{[(a - n)nx - \tan x]\sin nx}}{{{x^2}}} = 0,$ where $n$ is a non-zero real number,then $a$ is equal to

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