" Find the value of "k," so that the function "f(x)={[(log(1+ax)-log(1-bx))/(x),x!=0],[k,x=0]

f(x)={(log(1+2ax)-log(1-bx))/(x),x!=0x=0

Find the value of f(0) so that the function f(x)=([log (1+x//12)-log (1-x//8)])/x , x ne 0 is continuous on [0,8].

Find the value of k, so that the function is continuous at x=0f(x)={(1-cos4x),x!=0;k,x=0

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

If the function f(x) defined by f(x)=(log(1+3x)-log(1-2x))/(x),x!=0 and k

The value of f(0) so that the function f(x) = (log(sec^2 x))/(x sin x), x != 0 , is continuous at x = 0 is

The value of f(0) so that the function f(x) = (log(sec^2 x))/(x sin x), x != 0 , is continuous at x = 0 is

The value of f(0) so that f(x)=(log(1+x//a)-log(1-x//b))/(x) is continuous at x=0 is

If the function f(x) defined by f(x)= (log(1+3x)-"log"(1-2x))/x , x!=0 and k , x=0. Find k.

If the function f(x) defined by f(x)={((log(1+ax)-log(1-bx))/(k)", if "x!=0),(" k, if "x=0):} is continuous at x=0 , then find the value of k.