Question

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

• A. 0
• B. $\frac{n + 1}{n}$
• C. n
• D. $n + \frac{ 1}{n}$

Answer 1

Answer: (D)

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

Answer 2

Given, $lim_{x \to 0 } \frac{ \\{(a - n) nx - tan \, x \\} sin \, nx }{x^2} = 0$
\Rightarrow lim_{ x \to 0 } \Bigg \\{ (a - n) n - \frac{ tan \, x}{x} \Bigg \\} \frac{ sin \, n \, x}{ n \, x} \times n = 0
$\Rightarrow$ \hspace12mm { (a - n) n - 1 } n = 0
$\Rightarrow$ \hspace12mm (a - n ) n = 1 $\Rightarrow a = n + \frac{1}{n}$.