Section 5 e 與自然對數
5.1 lim
n→∞(1 −
n1)
n=lim
n→∞{
(1+ 11n−1)n−1
×
(1+11n−1)
}=
1e=e
−15.2 (1)ln
ee=log
ee=1 (2)e
lnx=x
lne=x
1=x (3)log
ax=
loglogexea
=
lnxlna5.3 (1)lim
x→1λ
lnx+lnλ
λx−1
=lim
λx→1lnλxλx−1−ln1=1
(2)lim
x→0lnsecxtan2x=lim
x→0seclnsecx2x−1=lim
x→0{
lnsecxsecx−ln1−1×
secx+11}=lim
secx→1{
lnsecxsecx−ln1−1×
secx+11}=
125.4 (1)lim
x→∞(1 +
λx)
x=lim
xλ→±∞
(1 +
1x λ)
x λ×λ
=e
λ(2)
同(1)
注意
