from the definition of eulers number we have
lim  (1+y)^(1/y)=e
y->0

so from this both the numerator and denomonator
tend to zero in the limit so we can use
l'hospital's rule to get
     (1+y)^(1/y)-e
lim  -------------=
y->0       y

lim  [d/dy (1+y)^(1/y)-e]=
y->0

lim  d/dy e^[ln(y+1)/y]=
y->0

lim (y+1)^(1/y)*(y-(y+1)ln(y+1))/[y^2(y+1)]
y->0

now we know that
lim (y+1)^(1/y)=e
y->0

so we have

lim  (y+1)^(1/y)*(y-(y+1)ln(y+1))/[y^2(y+1)]=
y->0

e*lim (y-(y+1)ln(y+1))/(y^2(y+1))
  y->0

and again we can use l'hospital's rule to get

e*lim -ln(y+1)/(y(3y+2))
   y->0

and using l'hospital on last time we get

e*lim -1/[(y+1)*(6y+2)]
   y->0

which can easily be seen to be -e/2
which agrees with numerical approximations that I got
on my calculator.