_{[x ][0 ]} [ f(1+x)/f(1) ]^1/x

= _{[x ][0 ]} [ f(1+x)/3 ]^1/x

 

= e[x ][0 ][ ((f(1+x)/3) - 1 )/x ]                 

  because when _{[x ][0 ]} f(x)^g(x) ,where _{[x ][0 ]} f(x)=1, _{[ x][0 ]} g(x) ^_>

 there [x ][0 ] f(x)g(x) = e [x ][0 ] (f(x) -1)/g(x)

 

=  e[x ][0 ][ f'(1+x) /3 ]         (using l hospital rule.....)

=  e²

 

I hope the answer is correct....plz rate....I haven't got one since long.....