If $\lim_{{x o 0}} \frac{{\sin(nx) [(a-n)nx - an x]}}{x^2} = 0$ $(n 0)$ then the value of 'a' is equal to

Question

If $\lim_{{x 	o 0}} \frac{{\sin(nx) [(a-n)nx - 	an x]}}{x^2} = 0$ $(n > 0)$ then the value of 'a' is equal to

Aicharya · Accepted Answer

$\lim_{{x 	o 0}} \frac{{\sin(nx)[(a-n)nx - 	an x]}}{{x^2}} = 0 \quad (n > 0)$ $\Rightarrow \lim_{{x 	o 0}} \left\{ (a-n)n - \frac{{	an x}}{{x}} ight\} \frac{{\sin nx}}{{nx}} 	imes n = 0$ $\Rightarrow (a-n)n - 1 = 0$ $\Rightarrow a - n = $\frac{1}{n}$ \Rightarrow a = n + \frac{1}{n}$

If $\lim_{{x \to 0}} \frac{{\sin(nx) [(a-n)nx - \tan x]}}{x^2} = 0$ $(n > 0)$ then the value of 'a' is equal…

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