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Question: If \(\lim_{x \rightarrow 0}\frac{\lbrack(a - n)nx - \tan x\rbrack\sin nx}{x^{2}} = 0,\) where n is n...

If lim⁡x→0[(a−n)nx−tan⁡x]sin⁡nxx2=0,\lim_{x \rightarrow 0}\frac{\lbrack(a - n)nx - \tan x\rbrack\sin nx}{x^{2}} = 0, where n is non-zero real number, then a is equal to

A

0

B

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

C

n

D

n+1nn + \frac{1}{n}

Answer

n+1nn + \frac{1}{n}

Explanation

Solution

lim⁡x→0nsin⁡nxnx.lim⁡x→0((a−n)n−tan⁡xx)=0\lim_{x \rightarrow 0}n\frac{\sin nx}{nx}.\lim_{x \rightarrow 0}\left( (a - n)n - \frac{\tan x}{x} \right) = 0

⇒ n[(a−n)n−1]=0⇒(a−n)n=1⇒a=n+1nn\lbrack(a - n)n - 1\rbrack = 0 \Rightarrow (a - n)n = 1 \Rightarrow a = n + \frac{1}{n}.