A) 0
B) \[\frac{n+1}{n}\]
C) n
D) \[n+\frac{1}{n}\]
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,n\frac{\sin nx}{nx}.\underset{x\to 0}{\mathop{\lim }}\,\left( (a-n)n-\frac{\tan x}{x} \right)=0\] Þ \[n((a-n)n-1)=0\Rightarrow (a-n)n=1\Rightarrow a=n+\frac{1}{n}\].
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