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.

Posted by Daniel
on 20090728 13:02:43 