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

Question

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

Dr. Abubacker · Accepted Answer

(d)
Given, $\lim_{x ightarrow 0}\frac{\left\{ (a - n)nx - 	an x ight\}\sin{nx}}{x^{2}} = 0$
$\Rightarrow \ \ \ \ \ \ \lim_{x ightarrow 0}{\left( (a - n)n - \frac{	an x}{x} ight).}\frac{\sin{nx}}{x} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \left\lbrack \left\{ a - n ight\} n - 1 ightbrack n = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ a = n + \frac{1}{n}$

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

Explanation

Reviewed by Dr. Abubacker

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