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$f(x)=\large\frac{\log(1+ax)-\log(1-bx)}{x}$ is not defined at $x=0$ .The value which should be assigned to f at $x=0$.So that it is continuous at $x=0$ is

$(a)\;a-b\qquad(b)\;a+b\qquad(c)\;\log a+\log b\qquad(d)\;\large\frac{a+b}{2}$

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$\lim\limits_{x\to 0}\large\frac{\log(1+ax)-\log(1-bx)}{x}\qquad \big(\frac{0}{0}\big)$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\Large\frac{1}{1+ax}.\normalsize a-\Large\frac{1}{1-bx}.\normalsize (-b)}{1}$
$\Rightarrow a+ b$
$f(0)=a+b$
Hence (b) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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