两个重要极限：
①
lim
x
→
0
sin
x
x
=
1
\lim_{x \to 0}\frac{\sin x}{x} = 1
x→0limxsinx=1
②
lim
x
→
∞
(
1
+
1
x
)
x
=
e
\lim_{x \to \infty}(1 + \frac{1}{x})^x = e
x→∞lim(1+x1)x=e
关于重要极限①的推导极限可以参考： 无穷小的等价代换
关于重要极限②的由来可以参考：自然常数e与重要极限
由重要极限②可以推导出：
lim
x
→
∞
(
1
+
1
x
)
x
⇒
lim
x
→
0
(
1
+
x
)
1
x
=
e
\lim_{x \to \infty}(1 + \frac{1}{x})^x \Rightarrow \lim_{x \to 0}(1 + x)^{\frac{1}{x}} = e
x→∞lim(1+x1)x⇒x→0lim(1+x)x1=e
重要极限②的例题
例：求
lim
x
→
∞
(
1
+
2
x
+
3
)
x
\lim_{x \to \infty} (1 + \frac{2}{x + 3})^x
limx→∞(1+x+32)x
解：
lim
x
→
∞
(
1
+
2
x
+
3
)
x
\lim_{x \to \infty} (1 + \frac{2}{x + 3})^x
x→∞lim(1+x+32)x
=
lim
x
→
∞
(
1
+
1
x
+
3
2
)
(
x
+
3
2
)
⋅
2
−
3
= \lim_{x \to \infty} (1 + \frac{1}{\frac{x + 3}{2}})^{(\frac{x+3}{2})·2 - 3}
=x→∞lim(1+2x+31)(2x+3)⋅2−3
=
lim
x
→
∞
(
1
+
1
x
+
3
2
)
x
+
3
2
⋅
2
⋅
(
1
+
1
x
+
3
2
)
−
3
= \lim_{x \to \infty} (1 + \frac{1}{\frac{x + 3}{2}})^{\frac{x+3}{2}·2} ·(1 + \frac{1}{\frac{x + 3}{2}})^{-3}
=x→∞lim(1+2x+31)2x+3⋅2⋅(1+2x+31)−3
=
lim
x
→
∞
(
1
+
1
x
+
3
2
)
x
+
3
2
⋅
2
=\lim_{x \to \infty} (1 + \frac{1}{\frac{x + 3}{2}})^{\frac{x+3}{2}·2}
=x→∞lim(1+2x+31)2x+3⋅2
=
e
2
= e^2
=e2
关于重要极限②，还有一个变体例题，很容易迷惑人，但实际上并不是用重要极限的方法来做
例：求
lim
x
→
∞
(
1
+
2
x
)
1
x
\lim_{x \to \infty}(1+2x)^{\frac{1}{x}}
x→∞lim(1+2x)x1
解：
(
1
+
2
x
)
1
x
=
e
ln
(
1
+
2
x
)
1
x
=
e
ln
(
1
+
2
x
)
x
(1+2x)^{\frac{1}{x}} = e^{\ln(1+2x)^{\frac{1}{x}}} = e^{\frac{\ln (1+2x)}{x}}
(1+2x)x1=eln(1+2x)x1=exln(1+2x)
所以：
lim
x
→
∞
(
1
+
2
x
)
1
x
=
lim
x
→
∞
e
ln
(
1
+
2
x
)
x
=
e
lim
x
→
∞
ln
(
1
+
2
x
)
x
\lim_{x \to \infty}(1+2x)^{\frac{1}{x}} = \lim_{x \to \infty}e^{\frac{\ln (1+2x)}{x}} = e^{\lim_{x \to \infty} \frac{\ln (1+2x)}{x}}
x→∞lim(1+2x)x1=x→∞limexln(1+2x)=elimx→∞xln(1+2x)
应用洛必达法则，得：
e
lim
x
→
∞
ln
(
1
+
2
x
)
x
=
e
lim
x
→
∞
2
1
+
2
x
=
e
0
=
1
e^{\lim_{x \to \infty} \frac{\ln (1+2x)}{x}} = e^{\lim_{x \to \infty} \frac{2}{1+2x}} = e^0 = 1
elimx→∞xln(1+2x)=elimx→∞1+2x2=e0=1
