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If $\lim\limits_{x\to 0}\large\frac{\log(3+x)-\log(3-x)}{x}$$=K$ the value of $K$ is

$(a)\;\large\frac{-2}{3}\qquad$$(b)\;0\qquad(c)\;\large\frac{-1}{3}\qquad$$(d)\;\large\frac{2}{3}$

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$\lim\limits_{x\to 0}\large\frac{\log(3+x)-\log(3-x)}{x}$$=K$(by 'L' Hospital rule)
$\lim\limits_{x\to 0}\large\frac{\Large\frac{1}{3+x}-\frac{-1}{3-x}}{1}$$=K$
$\Rightarrow K=\large\frac{2}{3}$
Hence (d) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 
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