6.)
u know ,
[x]
[a] [f(x)]
g(x) = e
lim [f(x) - 1] *g(x) as x->a ;
if g(x)->infinity , f(x)->1 as x->a ;
so,given = elim {(a^x+b^x+c^x - 3) / 3} * (2/x) ;
= e{(2/3) * lim {(a^x-1)+(b^x-1)+(c^x -1) / x} ;
now , lim as x->0 (a^x-1)/x = log a ;
So, given lim as x->0 = e
(2/3) * log(abc) = (abc)
2/3............ans.
