Question

If $\displaystyle \lim_{x \to 0} \frac{((a-n)nx - \tan x ) \sin nx}{x^2} = 0,$ where n is nonzero 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)

$\underset{\text{x} \rightarrow 0}{\text{lim}}\frac{\left(\right. \left(\right. a - n \left.\right) n \text{x} - \text{tan x)sin } \text{nx}}{\left(\text{x}\right)^{2}}=0$
$\Rightarrow \underset{\text{x} \rightarrow 0}{\text{lim}} \left(\left(\text{a} - \text{n}\right) \text{n} - \frac{\text{tan } \text{x}}{\text{x}}\right) \frac{\text{sin } \text{nx}}{\text{x}} = 0$
$\Rightarrow \left(\left(\text{a} - \text{n}\right) \text{n} - 1\right) \text{n} = 0$
$\Rightarrow \left(\text{a} - \text{n}\right) \text{n} - 1 = 0$ $\left(\because \text{n} \neq 0\right)$
$\therefore \text{a} = \text{n} + \frac{1}{\text{n}}$