If f(x)=log1+ax-log 1-bxx, x≠0,f(x) is continuous at x=0, then f(0)=

f0=limx→0log1+ax-log 1-bxx=limx→0log1+axx-limx→0log 1-bxx=a+b

If f(x)=log1+ax-log 1-bxx, x≠0,f(x) is continuous at x=0, then f(0)=

f0=limx→0log1+ax-log 1-bxx=limx→0log1+axx-limx→0log 1-bxx=a+b

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

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$a - b$

$\frac{a}{b}$

$a + b$

$\log a + \log b$

answer is B.

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Detailed Solution

$f (0) = \lim_{x \to 0} \frac{\log (1 + ax) - \log (1 - bx)}{x} = \lim_{x \to 0} \frac{\log (1 + ax)}{x} - \lim_{x \to 0} \frac{\log (1 - bx)}{x} = a + b$

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If f(x)=log1+ax-log 1-bxx, x≠0,f(x) is continuous at x=0, then f(0)=

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