The value of f(0) so that f(x)=(1+3x)1x is continuous at x=0 is

Ltx→0f(x)=f(0)                                   limx→cf(x)g(x)   → 1∞  then  limx→cf(x)g(x)=elimx→cg(x){f(x)-1}           =limx→0(1+3x)1x=  elimx→01x(1+3x-1)                            ⇒f(0)=e3