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根据换底公式
log
a
e
=
ln
(
e
)
ln
(
a
)
=
1
ln
(
a
)
\log_{a}{e}=\frac{\ln{(e)}}{\ln(a)}=\frac{1}{\ln(a)}
logae=ln(a)ln(e)=ln(a)1
lim
x
→
0
log
a
(
1
+
x
)
x
\lim\limits_{x\to{0}}{\frac{\log_a(1+x)}{x}}
x→0limxloga(1+x)=
lim
x
→
0
1
x
log
a
(
1
+
x
)
\lim\limits_{x\to 0}\frac{1}{x}{\log_a{(1+x)}}
x→0limx1loga(1+x)=
lim
x
→
0
log
a
(
(
1
+
x
)
1
x
)
=
log
a
(
e
)
=
1
ln
(
a
)
\lim\limits_{x\to{0}}\log_a((1+x)^{\frac{1}{x}})=\log_a(e)=\frac{1}{\ln(a)}
x→0limloga((1+x)x1)=loga(e)=ln(a)1 所以:
log
a
(
1
+
x
)
∼
1
ln
(
a
)
x
\log_a(1+x)\sim \frac{1}{\ln(a)}x
loga(1+x)∼ln(a)1x
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