If the function `f(x)=(log(1+ax)-log(1-bx))/(x), x ne 0` is continuous at x=0 then, f(0)= . . .

Options

A. loga-logb
B. a+b
C. loga+logb
D. a-b

Answer

B

Explanation

Given function
`f(x)=(log(1+ax)-log(1-bx))/(x), x ne 0`
is continuous at x=0
`therefore underset(xto0)(lim)f(x)=f(0)` . . . (i)
`underset(x to 0)(lim)f(x)=underset(x to 0)(lim)(log(1+ax)-log(1-bx))/(x)((0)/(0)"form")`
`=underset(x to 0)(lim)((a)/(1+ax)-((-b))/(1-bx))/(1)` (Using L' hospital's rule)
`=underset(x to 0)(lim)((a)/(1+bx)+(b)/(1-bx))=a+b`
From eq. (i) we get
`f(0)=a+b`