[" Given "f(x)=[log_(a)(a|[x]+[-x]|)^(x)|(a^(((|x|+-x|)/(|x|))))/((1)/(3+a^(|x)))|quad " for "|x|!=0;a>1],[0]quad " for "x=0

Given f(x)=[log_a (a|[x]+[-x]|)^x((a^((2/(([x]+[-x])))/(|x|)-5))/(3+a^(1/(|x|))) for |x| !=0 ; a lt 1 and 0 for x=0 where [ ] represents the integral part function, then

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

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

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

If f(x) = log_e ((1-x)/(1+x)) , then f' (0) is

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

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

If f(x) (2^(x)-1)/(1-3^(x)) , x != 0 is continuous at x = 0 then : f(0) =

If (x)={x,x 0 and g(x)=f(x)+|x| then lim_(x rarr0^(+))(log_(|sin x|)x)^(g(x))=