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

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(B) For the function $f(x)$ to be continuous at $x = 0$,the value $f(0)$ must be equal to $\lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\log(1 + ax) - \lo

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$a + b$

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(B) For the function $f(x)$ to be continuous at $x = 0$,the value $f(0)$ must be equal to $\lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$.Using the standard limit $\lim_{x 	o 0} \frac{\log(1 + kx)}{x} = k$,we can rewrite the limit as:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left( \frac{\log(1 + ax)}{x} - \frac{\log(1 - bx)}{x} ight)$$= \lim_{x 	o 0} \left( a \cdot \frac{\log(1 + ax)}{ax} - (-b) \cdot \frac{\log(1 - bx)}{-bx} ight)$$= a(1) + b(1) = a + b$.Therefore,$f(0) = a + b$.

The function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$ 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:

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