(B) For the function to be continuous at $x=0$,we must have $f(0) = \lim_{x \to 0} f(x)$.Using the standard limit $\lim_{x \to 0} \frac{\log(1+kx)}{x}

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(B) For the function to be continuous at $x=0$,we must have $f(0) = \lim_{x \to 0} f(x)$.Using the standard limit $\lim_{x \to 0} \frac{\log(1+kx)}{x} = k$,we have:$\lim_{x \to 0} \frac{\log(1+ax) - \log(1-bx)}{x} = \lim_{x \to 0} \left( \frac{\log(1+ax)}{x} - \frac{\log(1-bx)}{x} \right)$$= \lim_{x \to 0} \frac{\log(1+ax)}{x} - \lim_{x \to 0} \frac{\log(1-bx)}{x}$$= a - (-b) = a + b$.Thus,the value to be assigned is $a+b$.

Class 12 Mathematics Continuity and Differentiation Continuity

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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

KCET 2009 All KCET

A
$a-b$
B
$a+b$
C
$\log a + \log b$
D
$0$

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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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