maths (limit問題)急!!

The given limit is in the form 0/0

lim e^x -ex / (x - 1)^2
= lim e^x -e / 2(x - 1) (L'Hopital's Rule)
= lim e^x/2 (L'Hopital's Rule)
= e/2 = 1.359140914 = 1.36

2010-10-18 16:51:16 補充：
or,alternatively
e^x = (e)(e^x-1)
= e[1 + (x - 1) + (x - 1)^2/2! + (x - 1)^3/3!+...]
So lim (e^x - ex)/(x - 1)^2
= lim e[(x - 1)^2/2! + (x - 1)^3/3! + (x - 1)^4/4! + ...]/(x - 1)^2
= lim e/2! + e[(x - 1)/3! + (x - 1)^2/4! + ...]
= e/2
= 1.36

有無學過e^X=1+x+x^2/2!+x^3/3!+…?