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

$(a)\;0\qquad(b)\;\large\frac{n+1}{n}$$\qquad(c)\;n\qquad(d)\;n+\large\frac{1}{n}$

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$\lim\limits_{x\to 0}\large\frac{n\sin nx}{nx}.$$\lim\limits_{x\to 0}((a-x)n-\large\frac{\tan x}{n})$$=0$
$\Rightarrow n((a-n)n-1)=0$
$\Rightarrow (a-n)^n=1$
$\Rightarrow a^n=1+n^2$
$\Rightarrow a=\large\frac{1}{n}+$$n$
Hence (d) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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