If the function f(x) = frac{log(1 + ax) - log | vedclass

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(B) Given that the function $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.$f(0) = \lim_{x 	o 0} \frac{\log(1 + ax) - \log(1 -

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$a + b$

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(B) Given that the function $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.$f(0) = \lim_{x 	o 0} \frac{\log(1 + ax) - \log(1 - bx)}{x}$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:$f(0) = \lim_{x 	o 0} \frac{\frac{d}{dx}(\log(1 + ax)) - \frac{d}{dx}(\log(1 - bx))}{\frac{d}{dx}(x)}$.$f(0) = \lim_{x 	o 0} \frac{\frac{a}{1 + ax} - \frac{-b}{1 - bx}}{1}$.$f(0) = \lim_{x 	o 0} \left( \frac{a}{1 + ax} + \frac{b}{1 - bx} ight)$.Substituting $x = 0$:$f(0) = \frac{a}{1 + 0} + \frac{b}{1 - 0} = a + b$.

If the function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$,$x \neq 0$ is continuous at $x = 0$,then $f(0) = $ . . . . . .

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