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

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

A
0
B
$\frac{n}{n + 1}$
C
$n$
D
$n + (1 / n)$

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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