If f(x)=((a−n)nx−tan⁡x)sin⁡nxx2,x≠0and limx→0 f(x)=0,n is a non-zero real number, then a isequal to

limx→0 f(x)=0⇒ limx→0 sin⁡nxnx×(n)(a−n)n−tan⁡xx=0⇒ n((a−n)n−1)=0⇒ (a−n)n−1=0⇒a=n+1n

If f(x)=((a−n)nx−tan⁡x)sin⁡nxx2,x≠0and limx→0 f(x)=0,n is a non-zero real number, then a isequal to

limx→0 f(x)=0⇒ limx→0 sin⁡nxnx×(n)(a−n)n−tan⁡xx=0⇒ n((a−n)n−1)=0⇒ (a−n)n−1=0⇒a=n+1n

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If $f (x) = \frac{((a - n) n x - \tan x) \sin n x}{x^{2}}, x \neq 0$
and $lim_{x \to 0} f (x) = 0, n$ is $a$ non-zero real number, then $a$ is
equal to

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$n$

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

$n + (1 / n)$

answer is D.

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Detailed Solution

$lim_{x \to 0} f (x) = 0$

$\begin{array}{ll} \Rightarrow lim_{x \to 0} \frac{\sin n x}{n x} \times (n) [(a - n) n - \frac{\tan x}{x}] = 0 \\ \Rightarrow n ((a - n) n - 1) = 0 \\ \Rightarrow (a - n) n - 1 = 0 \Rightarrow a = n + \frac{1}{n} \end{array}$

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If f(x)=((a−n)nx−tan⁡x)sin⁡nxx2,x≠0and limx→0 f(x)=0,n is a non-zero real number, then a isequal to

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If $f (x) = \frac{((a - n) n x - \tan x) \sin n x}{x^{2}}, x \neq 0$
and $lim_{x \to 0} f (x) = 0, n$ is $a$ non-zero real number, then $a$ is
equal to