(iv)f(x)={[(e^(x)-1)/(log(1+2x))," if "x!=0" at "x=0],[7," if "x=0]

f(x)={(e^(x)-1)/(log(1+2x)),, if x!=0quad 7,quad if x=0 at x=0

'f(x)={(e^(^^)x-1)/(log(1+2x)),|x!=0Kx=0at

If f(x) {:(=(log (1+3x )-log(1-2x))/x", if " x != 0 ),(=a", if " x =0 ):} is continuous at x = 0 , then a =

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

If f(x)=(e^(1//x)-1)/(e^(1//x)+1)" for "x ne 0, f(0)=0" then at x=0, f(x) is"

If f(x)={:{((e^(1/x)-1)/(e^(1/x)+1)", for " x !=0),(1", for " x=0):} , then f is

The function f(x)=x*e^(-((1)/(|x|)+(1)/(x))) if x!=0 and f(x)=0 if x=0 then

f(x)={(e^x-1)/(log(1+2x)),\ \ \ if\ x!=0\ \ \ \ \ \ \ \ \ 7,\ \ \ \ \ \ \ \ \ \ \ if\ x=0 at x=0

Find the points of discontinuity, if any, of the following function: f(x)={(e^x-1)/((log)_e(1+2x))\ \ \ ,\ \ \ if\ x!=0\ \ \ \ \ 7\ \ \ ,\ \ \ if\ x=0

f(x)=((2^(x)-1)^(2))/(tan x log(1+x)),x!=0 and f(x)=2log2,x=0 at x=0 is: