Замеча́тельные преде́лы — термины, использующиеся в советских и российских учебниках по математическому анализу для обозначения двух широко известных математических тождеств со взятием предела :
Первый замечательный предел:
lim
x
→ → -->
0
sin
-->
x
x
=
1.
{\displaystyle \lim _{x\to 0}{\frac {\sin x}{x}}=1.}
Второй замечательный предел:
lim
x
→ → -->
∞ ∞ -->
(
1
+
1
x
)
x
=
e
.
{\displaystyle \lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e.}
lim
x
→ → -->
0
sin
-->
x
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\sin x}{x}}=1}
Доказательство:
Рассмотрим односторонние пределы
lim
x
→ → -->
+
0
sin
-->
x
x
{\displaystyle \lim _{x\to +0}{\frac {\sin x}{x}}}
lim
x
→ → -->
− − -->
0
sin
-->
x
x
{\displaystyle \lim _{x\to {\displaystyle -}0}{\frac {\sin x}{x}}}
Рассмотрим случай
x
∈ ∈ -->
(
0
;
π π -->
2
)
{\displaystyle x\in \left(0;{\frac {\pi }{2}}\right)}
O
X
{\displaystyle OX}
K
{\displaystyle K}
L
{\displaystyle L}
A
=
(
1
;
0
)
{\displaystyle A=\left(1;0\right)}
H
{\displaystyle H}
K
{\displaystyle K}
O
X
{\displaystyle OX}
Очевидно, что:
S
△ △ -->
O
K
A
<
S
s
e
c
t
K
O
A
<
S
△ △ -->
O
A
L
{\displaystyle S_{\triangle OKA}<S_{sectKOA}<S_{\triangle OAL}}
(где
S
s
e
c
t
K
O
A
{\displaystyle S_{sectKOA}}
K
O
A
{\displaystyle KOA}
Поскольку
|
K
H
|
=
sin
-->
x
,
|
L
A
|
=
tg
-->
x
{\displaystyle \left|KH\right|=\sin x,\,\left|LA\right|=\operatorname {tg} x}
S
△ △ -->
O
K
A
=
1
2
⋅ ⋅ -->
|
O
A
|
⋅ ⋅ -->
|
K
H
|
=
1
2
⋅ ⋅ -->
1
⋅ ⋅ -->
sin
-->
x
=
sin
-->
x
2
{\displaystyle S_{\triangle OKA}={\frac {1}{2}}\cdot \left|OA\right|\cdot \left|KH\right|={\frac {1}{2}}\cdot 1\cdot \sin x={\frac {\sin x}{2}}}
S
s
e
c
t
K
O
A
=
1
2
⋅ ⋅ -->
|
O
A
|
2
⋅ ⋅ -->
x
=
x
2
{\displaystyle S_{sectKOA}={\frac {1}{2}}\cdot \left|OA\right|^{2}\cdot x={\frac {x}{2}}}
S
△ △ -->
O
A
L
=
1
2
⋅ ⋅ -->
|
O
A
|
⋅ ⋅ -->
|
L
A
|
=
tg
-->
x
2
{\displaystyle S_{\triangle OAL}={\frac {1}{2}}\cdot \left|OA\right|\cdot \left|LA\right|={\frac {\operatorname {tg} x}{2}}}
Подставляя в (1), получим:
sin
-->
x
2
<
x
2
<
tg
-->
x
2
{\displaystyle {\frac {\sin x}{2}}<{\frac {x}{2}}<{\frac {\operatorname {tg} x}{2}}}
Так как при
x
→ → -->
+
0
:
sin
-->
x
>
0
,
x
>
0
,
tg
-->
x
>
0
{\displaystyle x\to +0:\sin x>0,\,x>0,\,\operatorname {tg} x>0}
1
tg
-->
x
<
1
x
<
1
sin
-->
x
{\displaystyle {\frac {1}{\operatorname {tg} x}}<{\frac {1}{x}}<{\frac {1}{\sin x}}}
Умножаем на
sin
-->
x
{\displaystyle \sin x}
cos
-->
x
<
sin
-->
x
x
<
1
{\displaystyle \cos x<{\frac {\sin x}{x}}<1}
Перейдём к пределу:
lim
x
→ → -->
+
0
cos
-->
x
⩽ ⩽ -->
lim
x
→ → -->
+
0
sin
-->
x
x
⩽ ⩽ -->
1
{\displaystyle \lim _{x\to +0}\cos x\leqslant \lim _{x\to +0}{\frac {\sin x}{x}}\leqslant 1}
1
⩽ ⩽ -->
lim
x
→ → -->
+
0
sin
-->
x
x
⩽ ⩽ -->
1
{\displaystyle 1\leqslant \lim _{x\to +0}{\frac {\sin x}{x}}\leqslant 1}
lim
x
→ → -->
+
0
sin
-->
x
x
=
1
{\displaystyle \lim _{x\to +0}{\frac {\sin x}{x}}=1}
Найдём левый односторонний предел (так как функция четна, в этом нет необходимости, достаточно доказать это для правого предела):
lim
x
→ → -->
− − -->
0
sin
-->
x
x
=
[
u
=
− − -->
x
x
=
− − -->
u
u
→ → -->
+
0
x
→ → -->
− − -->
0
]
=
lim
u
→ → -->
+
0
sin
-->
(
− − -->
u
)
− − -->
u
=
lim
u
→ → -->
+
0
− − -->
sin
-->
(
u
)
− − -->
u
=
lim
u
→ → -->
+
0
sin
-->
(
u
)
u
=
1
{\displaystyle \lim _{x\to -0}{\frac {\sin x}{x}}=\left[{\begin{matrix}u=-x\\x=-u\\u\to +0\\x\to -0\end{matrix}}\right]=\lim _{u\to +0}{\frac {\sin(-u)}{-u}}=\lim _{u\to +0}{\frac {-\sin(u)}{-u}}=\lim _{u\to +0}{\frac {\sin(u)}{u}}=1}
Правый и левый односторонний пределы существуют и равны 1, а значит и сам предел равен 1.
Следствия:
lim
x
→ → -->
0
tg
-->
x
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\operatorname {tg} x}{x}}=1}
lim
x
→ → -->
0
arcsin
-->
x
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\operatorname {arcsin} x}{x}}=1}
lim
x
→ → -->
0
arctg
-->
x
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\operatorname {arctg} x}{x}}=1}
lim
x
→ → -->
0
1
− − -->
cos
-->
x
x
2
2
=
1
{\displaystyle \lim _{x\to 0}{\frac {1-\cos x}{\frac {x^{2}}{2}}}=1}
lim
x
→ → -->
0
tg
-->
x
x
=
lim
x
→ → -->
0
sin
-->
x
x
cos
-->
x
=
lim
x
→ → -->
0
sin
-->
x
x
⋅ ⋅ -->
lim
x
→ → -->
0
1
cos
-->
x
=
1
⋅ ⋅ -->
1
=
1
{\displaystyle \lim _{x\to 0}{\frac {\operatorname {tg} x}{x}}=\lim _{x\to 0}{\frac {\sin x}{x\cos x}}=\lim _{x\to 0}{\frac {\sin x}{x}}\cdot \lim _{x\to 0}{\frac {1}{\cos x}}=1\cdot 1=1}
lim
x
→ → -->
0
arcsin
-->
x
x
=
[
u
=
arcsin
-->
x
x
=
sin
-->
u
u
→ → -->
0
x
→ → -->
0
]
=
lim
u
→ → -->
0
u
sin
-->
u
=
1
{\displaystyle \lim _{x\to 0}{\frac {\operatorname {arcsin} x}{x}}=\left[{\begin{matrix}u=\operatorname {arcsin} x\\x=\sin u\\u\to 0\\x\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {u}{\sin u}}=1}
lim
x
→ → -->
0
arctg
-->
x
x
=
[
u
=
arctg
-->
x
x
=
tg
-->
u
u
→ → -->
0
x
→ → -->
0
]
=
lim
u
→ → -->
0
u
tg
-->
u
=
1
{\displaystyle \lim _{x\to 0}{\frac {\operatorname {arctg} x}{x}}=\left[{\begin{matrix}u=\operatorname {arctg} x\\x=\operatorname {tg} u\\u\to 0\\x\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {u}{\operatorname {tg} u}}=1}
lim
x
→ → -->
0
1
− − -->
cos
-->
x
x
2
2
=
lim
x
→ → -->
0
2
⋅ ⋅ -->
sin
2
-->
x
2
2
⋅ ⋅ -->
(
x
2
)
2
=
lim
x
2
→ → -->
0
(
sin
-->
x
2
x
2
)
2
=
1
2
=
1
{\displaystyle \lim _{x\to 0}{\frac {1-\cos x}{\frac {x^{2}}{2}}}=\lim _{x\to 0}{\frac {2\cdot \sin ^{2}{\frac {x}{2}}}{2\cdot \left({\frac {x}{2}}\right)^{2}}}=\lim _{{\frac {x}{2}}\to 0}\left({\frac {\sin {\frac {x}{2}}}{\frac {x}{2}}}\right)^{2}=1^{2}=1}
lim
x
→ → -->
∞ ∞ -->
(
1
+
1
x
)
x
=
e
{\displaystyle \lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e}
Доказательство существования второго замечательного предела:
Доказательство для натуральных значений x
◂ ◂ -->
{\displaystyle \blacktriangleleft }
x
n
=
(
1
+
1
n
)
n
;
n
∈ ∈ -->
N
{\displaystyle x_{n}=\left(1+{\frac {1}{n}}\right)^{n};\ n\in \mathbb {N} }
По формуле бинома Ньютона :
(
a
+
b
)
n
=
a
n
+
n
1
⋅ ⋅ -->
a
n
− − -->
1
⋅ ⋅ -->
b
+
n
(
n
− − -->
1
)
1
⋅ ⋅ -->
2
⋅ ⋅ -->
a
n
− − -->
2
⋅ ⋅ -->
b
2
+
.
.
.
+
n
(
n
− − -->
1
)
(
n
− − -->
2
)
.
.
.
(
n
− − -->
(
n
− − -->
1
)
)
1
⋅ ⋅ -->
2
⋅ ⋅ -->
3
⋅ ⋅ -->
.
.
.
⋅ ⋅ -->
n
⋅ ⋅ -->
b
n
;
n
∈ ∈ -->
N
{\displaystyle (a+b)^{n}=a^{n}~+~{\frac {n}{1}}\cdot a^{n-1}\cdot b~+~{\frac {n(n-1)}{1\cdot 2}}\cdot a^{n-2}\cdot b^{2}~+~...~+~{\frac {n(n-1)(n-2)...(n-(n-1))}{1\cdot 2\cdot 3\cdot ...\cdot n}}\cdot b^{n};\ n\in \mathbb {N} }
Полагая
a
=
1
;
b
=
1
n
{\displaystyle a=1;~b={\frac {1}{n}}}
(
1
+
1
n
)
n
=
1
+
n
1
⋅ ⋅ -->
1
n
+
n
(
n
− − -->
1
)
1
⋅ ⋅ -->
2
⋅ ⋅ -->
1
n
2
+
n
(
n
− − -->
1
)
(
n
− − -->
2
)
1
⋅ ⋅ -->
2
⋅ ⋅ -->
3
⋅ ⋅ -->
1
n
3
+
.
.
.
+
n
(
n
− − -->
1
)
(
n
− − -->
2
)
.
.
.
(
n
− − -->
(
n
− − -->
1
)
)
1
⋅ ⋅ -->
2
⋅ ⋅ -->
3
⋅ ⋅ -->
.
.
.
⋅ ⋅ -->
n
⋅ ⋅ -->
1
n
n
=
{\displaystyle \left(1+{\frac {1}{n}}\right)^{n}=1~+~{\frac {n}{1}}\cdot {\frac {1}{n}}~+~{\frac {n(n-1)}{1\cdot 2}}\cdot {\frac {1}{n^{2}}}~+~{\frac {n(n-1)(n-2)}{1\cdot 2\cdot 3}}\cdot {\frac {1}{n^{3}}}~+~...~+~{\frac {n(n-1)(n-2)...(n-(n-1))}{1\cdot 2\cdot 3\cdot ...\cdot n}}\cdot {\frac {1}{n^{n}}}=}
=
1
+
1
+
1
1
⋅ ⋅ -->
2
⋅ ⋅ -->
(
1
− − -->
1
n
)
+
1
1
⋅ ⋅ -->
2
⋅ ⋅ -->
3
⋅ ⋅ -->
(
1
− − -->
1
n
)
⋅ ⋅ -->
(
1
− − -->
2
n
)
+
.
.
.
+
1
1
⋅ ⋅ -->
2
⋅ ⋅ -->
3
⋅ ⋅ -->
.
.
.
⋅ ⋅ -->
n
⋅ ⋅ -->
(
1
− − -->
1
n
)
⋅ ⋅ -->
(
1
− − -->
2
n
)
⋅ ⋅ -->
.
.
.
⋅ ⋅ -->
(
1
− − -->
n
− − -->
1
n
)
{\displaystyle =1~+~1~+~{\frac {1}{1\cdot 2}}\cdot \left(1-{\frac {1}{n}}\right)~+~{\frac {1}{1\cdot 2\cdot 3}}\cdot \left(1-{\frac {1}{n}}\right)\cdot \left(1-{\frac {2}{n}}\right)~+~...~+~{\frac {1}{1\cdot 2\cdot 3\cdot ...\cdot n}}\cdot \left(1-{\frac {1}{n}}\right)\cdot \left(1-{\frac {2}{n}}\right)\cdot ...\cdot \left(1-{\frac {n-1}{n}}\right)}
С увеличением
n
{\displaystyle n}
n
{\displaystyle n}
1
n
{\displaystyle {\frac {1}{n}}}
(
1
− − -->
1
n
)
,
(
1
− − -->
2
n
)
,
.
.
.
{\displaystyle \left(1-{\frac {1}{n}}\right),\left(1-{\frac {2}{n}}\right),...}
{
x
n
}
=
{
(
1
+
1
n
)
n
}
;
n
∈ ∈ -->
N
{\displaystyle \{x_{n}\}=\left\{\left(1+{\frac {1}{n}}\right)^{n}\right\};\ n\in \mathbb {N} }
возрастающая , при этом
(
1
+
1
n
)
n
≥ ≥ -->
2
,
n
∈ ∈ -->
N
{\displaystyle \left(1+{\frac {1}{n}}\right)^{n}\geq 2,n\in \mathbb {N} }
Покажем, что она ограничена. Заменим каждую скобку в правой части равенства на единицу, правая часть увеличится, получим неравенство
(
1
+
1
n
)
n
<
1
+
1
+
1
1
⋅ ⋅ -->
2
+
1
1
⋅ ⋅ -->
2
⋅ ⋅ -->
3
+
.
.
.
+
1
1
⋅ ⋅ -->
2
⋅ ⋅ -->
3
⋅ ⋅ -->
.
.
.
⋅ ⋅ -->
n
{\displaystyle \left(1+{\frac {1}{n}}\right)^{n}<1+1+{\frac {1}{1\cdot 2}}+{\frac {1}{1\cdot 2\cdot 3}}~+~...~+~{\frac {1}{1\cdot 2\cdot 3\cdot ...\cdot n}}}
Усилим полученное неравенство, заменим 3,4,5, …, стоящие в знаменателях дробей, числом 2:
(
1
+
1
n
)
n
<
1
+
(
1
+
1
2
+
1
2
2
+
.
.
.
+
1
2
n
− − -->
1
)
{\displaystyle \left(1+{\frac {1}{n}}\right)^{n}<1+\left(1+{\frac {1}{2}}+{\frac {1}{2^{2}}}+...+{\frac {1}{2^{n-1}}}\right)}
Сумму в скобке найдём по формуле суммы членов геометрической прогрессии:
1
+
1
2
+
1
2
2
+
.
.
.
+
1
2
n
− − -->
1
=
1
⋅ ⋅ -->
(
1
− − -->
(
1
2
)
n
)
1
− − -->
1
2
=
2
⋅ ⋅ -->
(
1
− − -->
1
2
n
)
<
2
{\displaystyle 1+{\frac {1}{2}}+{\frac {1}{2^{2}}}+...+{\frac {1}{2^{n-1}}}={\frac {1\cdot \left(1-({\frac {1}{2}})^{n}\right)}{1-{\frac {1}{2}}}}=2\cdot \left(1-{\frac {1}{2^{n}}}\right)<2}
Поэтому
(
1
+
1
n
)
n
<
1
+
2
=
3
{\displaystyle \left(1+{\frac {1}{n}}\right)^{n}<1+2=3}
Итак, последовательность ограничена сверху, при этом
∀ ∀ -->
n
∈ ∈ -->
N
{\displaystyle \forall n\in \mathbb {N} }
2
≤ ≤ -->
(
1
+
1
n
)
n
<
3
{\displaystyle 2\leq \left(1+{\frac {1}{n}}\right)^{n}<3}
Следовательно, на основании теоремы Вейерштрасса (критерий сходимости последовательности) последовательность
x
n
=
(
1
+
1
n
)
n
,
n
∈ ∈ -->
N
{\displaystyle x_{n}=\left(1+{\frac {1}{n}}\right)^{n},\ n\in \mathbb {N} }
e
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
)
n
=
e
{\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e}
▸ ▸ -->
{\displaystyle \blacktriangleright }
◂ ◂ -->
{\displaystyle \blacktriangleleft }
lim
x
→ → -->
∞ ∞ -->
(
1
+
1
x
)
x
=
e
;
x
∈ ∈ -->
R
{\displaystyle \lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e;\ x\in \mathbb {R} }
1. Пусть
x
→ → -->
+
∞ ∞ -->
{\displaystyle x\rightarrow +\infty }
n
⩽ ⩽ -->
x
<
n
+
1
{\displaystyle n\leqslant x<n+1}
n
=
[
x
]
{\displaystyle n=[x]}
Отсюда следует:
1
n
+
1
<
1
x
⩽ ⩽ -->
1
n
⟺ ⟺ -->
1
+
1
n
+
1
<
1
+
1
x
⩽ ⩽ -->
1
+
1
n
{\displaystyle {\frac {1}{n+1}}<{\frac {1}{x}}\leqslant {\frac {1}{n}}~~\Longleftrightarrow ~~1+{\frac {1}{n+1}}<1+{\frac {1}{x}}\leqslant 1+{\frac {1}{n}}}
(
1
+
1
n
+
1
)
n
<
(
1
+
1
x
)
x
⩽ ⩽ -->
(
1
+
1
n
)
n
+
1
{\displaystyle \left(1+{\frac {1}{n+1}}\right)^{n}<\left(1+{\frac {1}{x}}\right)^{x}\leqslant \left(1+{\frac {1}{n}}\right)^{n+1}}
Если
x
→ → -->
+
∞ ∞ -->
{\displaystyle x\rightarrow +\infty }
n
→ → -->
∞ ∞ -->
{\displaystyle n\rightarrow \infty }
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
)
n
=
e
{\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e}
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
+
1
)
n
=
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
+
1
)
n
+
1
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
+
1
)
=
e
1
=
e
{\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n+1}}\right)^{n}={\frac {\lim \limits _{n\to \infty }(1+{\frac {1}{n+1}})^{n+1}}{\lim \limits _{n\to \infty }\left(1+{\frac {1}{n+1}}\right)}}={\frac {e}{1}}=e}
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
)
n
+
1
=
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
)
n
⋅ ⋅ -->
lim
n
→ → -->
∞ ∞ -->
(
1
+
1
n
)
=
e
⋅ ⋅ -->
1
=
e
{\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n+1}=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}\cdot \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)=e\cdot 1=e}
По признаку (о пределе промежуточной функции) существования пределов
lim
x
→ → -->
+
∞ ∞ -->
(
1
+
1
x
)
x
=
e
{\displaystyle \lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e}
2 . Пусть
x
→ → -->
− − -->
∞ ∞ -->
{\displaystyle x\to -\infty }
− − -->
x
=
t
{\displaystyle -x=t}
lim
x
→ → -->
− − -->
∞ ∞ -->
(
1
+
1
x
)
x
=
lim
t
→ → -->
+
∞ ∞ -->
(
1
− − -->
1
t
)
− − -->
t
=
lim
t
→ → -->
+
∞ ∞ -->
(
t
t
− − -->
1
)
t
=
lim
t
→ → -->
+
∞ ∞ -->
(
1
+
1
t
− − -->
1
)
t
=
{\displaystyle \lim _{x\to -\infty }\left(1+{\frac {1}{x}}\right)^{x}=\lim _{t\to +\infty }\left(1-{\frac {1}{t}}\right)^{-t}=\lim _{t\to +\infty }\left({\frac {t}{t-1}}\right)^{t}=\lim _{t\to +\infty }\left(1+{\frac {1}{t-1}}\right)^{t}=}
=
lim
t
→ → -->
+
∞ ∞ -->
(
1
+
1
t
− − -->
1
)
t
− − -->
1
⋅ ⋅ -->
lim
t
→ → -->
+
∞ ∞ -->
(
1
+
1
t
− − -->
1
)
1
=
e
⋅ ⋅ -->
1
=
e
{\displaystyle =\lim _{t\to +\infty }\left(1+{\frac {1}{t-1}}\right)^{t-1}\cdot \lim _{t\to +\infty }\left(1+{\frac {1}{t-1}}\right)^{1}=e\cdot 1=e}
Очевидно, из двух этих случаев вытекает, что
lim
x
→ → -->
∞ ∞ -->
(
1
+
1
x
)
x
=
e
{\displaystyle \lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e}
▸ ▸ -->
{\displaystyle \blacktriangleright }
Следствия
lim
x
→ → -->
0
(
1
+
x
)
1
/
x
=
e
{\displaystyle \lim _{x\to 0}\left(1+x\right)^{1/x}=e}
lim
x
→ → -->
∞ ∞ -->
(
1
+
k
x
)
x
=
e
k
{\displaystyle \lim _{x\to \infty }\left(1+{\frac {k}{x}}\right)^{x}=e^{k}}
lim
x
→ → -->
0
ln
-->
(
1
+
x
)
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=1}
lim
x
→ → -->
0
e
x
− − -->
1
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}=1}
lim
x
→ → -->
0
a
x
− − -->
1
x
ln
-->
a
=
1
{\displaystyle \lim _{x\to 0}{\frac {a^{x}-1}{x\ln a}}=1}
a
>
0
{\displaystyle a>0}
a
≠ ≠ -->
1
{\displaystyle a\neq 1}
lim
x
→ → -->
0
(
1
+
x
)
α α -->
− − -->
1
α α -->
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {(1+x)^{\alpha }-1}{\alpha x}}=1}
lim
x
→ → -->
∞ ∞ -->
(
1
− − -->
k
x
)
x
=
e
− − -->
k
{\displaystyle \lim _{x\to \infty }\left(1-{\frac {k}{x}}\right)^{x}=e^{-k}}
lim
u
→ → -->
0
(
1
+
u
)
1
u
=
[
u
=
1
/
x
x
→ → -->
∞ ∞ -->
]
=
lim
x
→ → -->
∞ ∞ -->
(
1
+
1
x
)
x
=
e
{\displaystyle \lim _{u\to 0}(1+u)^{\frac {1}{u}}=\left[{\begin{matrix}u=1/x\\x\to \infty \end{matrix}}\right]=\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e}
lim
x
→ → -->
∞ ∞ -->
(
1
+
k
x
)
x
=
[
u
=
x
/
k
x
=
k
u
u
→ → -->
∞ ∞ -->
x
→ → -->
∞ ∞ -->
]
=
lim
u
→ → -->
∞ ∞ -->
(
1
+
1
u
)
k
u
=
(
lim
u
→ → -->
∞ ∞ -->
(
1
+
1
u
)
u
)
k
=
e
k
{\displaystyle \lim _{x\to \infty }\left(1+{\frac {k}{x}}\right)^{x}=\left[{\begin{matrix}u=x/k\\x=ku\\u\to \infty \\x\to \infty \end{matrix}}\right]=\lim _{u\to \infty }\left(1+{\frac {1}{u}}\right)^{ku}=\left(\lim _{u\to \infty }\left(1+{\frac {1}{u}}\right)^{u}\right)^{k}=e^{k}}
lim
x
→ → -->
0
ln
-->
(
1
+
x
)
x
=
lim
x
→ → -->
0
1
x
ln
-->
(
1
+
x
)
=
lim
x
→ → -->
0
ln
-->
(
(
1
+
x
)
1
x
)
=
ln
-->
e
=
1
{\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=\lim _{x\to 0}{\frac {1}{x}}\ln(1+x)=\lim _{x\to 0}\ln((1+x)^{\frac {1}{x}})=\ln e=1}
lim
x
→ → -->
0
e
x
− − -->
1
x
=
[
u
=
e
x
− − -->
1
x
=
ln
-->
(
1
+
u
)
x
→ → -->
0
u
→ → -->
0
]
=
lim
u
→ → -->
0
u
ln
-->
(
1
+
u
)
=
1
{\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}=\left[{\begin{matrix}u=e^{x}-1\\x=\ln(1+u)\\x\to 0\\u\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {u}{\ln(1+u)}}=1}
lim
x
→ → -->
0
a
x
− − -->
1
x
ln
-->
a
=
lim
x
→ → -->
0
e
ln
-->
(
a
x
)
− − -->
1
x
ln
-->
a
=
lim
x
→ → -->
0
e
x
ln
-->
a
− − -->
1
x
ln
-->
a
=
[
u
=
x
ln
-->
a
u
→ → -->
0
x
→ → -->
0
]
=
lim
u
→ → -->
0
e
u
− − -->
1
u
=
1
{\displaystyle \lim _{x\to 0}{\frac {a^{x}-1}{x\ln a}}=\lim _{x\to 0}{\frac {e^{\ln(a^{x})}-1}{x\ln a}}=\lim _{x\to 0}{\frac {e^{x\ln a}-1}{x\ln a}}=\left[{\begin{matrix}u=x\ln a\\u\to 0\\x\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {e^{u}-1}{u}}=1}
lim
x
→ → -->
0
(
1
+
x
)
α α -->
− − -->
1
α α -->
x
=
lim
x
→ → -->
0
e
α α -->
ln
-->
(
1
+
x
)
− − -->
1
α α -->
x
=
lim
x
→ → -->
0
e
α α -->
ln
-->
(
1
+
x
)
− − -->
1
α α -->
ln
-->
(
1
+
x
)
⋅ ⋅ -->
lim
x
→ → -->
0
ln
-->
(
1
+
x
)
x
=
{\displaystyle \lim _{x\to 0}{\frac {(1+x)^{\alpha }-1}{\alpha x}}=\lim _{x\to 0}{\frac {e^{\alpha \ln(1+x)}-1}{\alpha x}}=\lim _{x\to 0}{\frac {e^{\alpha \ln(1+x)}-1}{\alpha \ln(1+x)}}\cdot \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=}
=
lim
x
→ → -->
0
e
α α -->
ln
-->
(
1
+
x
)
− − -->
1
α α -->
ln
-->
(
1
+
x
)
⋅ ⋅ -->
1
=
[
u
=
α α -->
ln
-->
(
1
+
x
)
x
→ → -->
0
u
→ → -->
0
]
=
lim
u
→ → -->
0
e
u
− − -->
1
u
=
1
{\displaystyle =\lim _{x\to 0}{\frac {e^{\alpha \ln(1+x)}-1}{\alpha \ln(1+x)}}\cdot 1=\left[{\begin{matrix}u=\alpha \ln(1+x)\\x\to 0\\u\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {e^{u}-1}{u}}=1}
Замечательные пределы и их следствия используются при раскрытии неопределённостей для нахождения других пределов.
![{\displaystyle \lim _{x\to -0}{\frac {\sin x}{x}}=\left[{\begin{matrix}u=-x\\x=-u\\u\to +0\\x\to -0\end{matrix}}\right]=\lim _{u\to +0}{\frac {\sin(-u)}{-u}}=\lim _{u\to +0}{\frac {-\sin(u)}{-u}}=\lim _{u\to +0}{\frac {\sin(u)}{u}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb9f4662b6f0e8b77bf8f3fc6ca54c04d14acce2)
![{\displaystyle \lim _{x\to 0}{\frac {\operatorname {arcsin} x}{x}}=\left[{\begin{matrix}u=\operatorname {arcsin} x\\x=\sin u\\u\to 0\\x\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {u}{\sin u}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41264aec3d1a47592437838744da3293740e8a57)
![{\displaystyle \lim _{x\to 0}{\frac {\operatorname {arctg} x}{x}}=\left[{\begin{matrix}u=\operatorname {arctg} x\\x=\operatorname {tg} u\\u\to 0\\x\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {u}{\operatorname {tg} u}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c5a3f7b6e957799a9c3e2ae192cddbb346df644)
![{\displaystyle n=[x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4d7b9bfe1601e524b07a917892bf1a4c462b830)
![{\displaystyle \lim _{u\to 0}(1+u)^{\frac {1}{u}}=\left[{\begin{matrix}u=1/x\\x\to \infty \end{matrix}}\right]=\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8837fc0ee9e3a9b2b8a466920af3f1b52c2d75fb)
![{\displaystyle \lim _{x\to \infty }\left(1+{\frac {k}{x}}\right)^{x}=\left[{\begin{matrix}u=x/k\\x=ku\\u\to \infty \\x\to \infty \end{matrix}}\right]=\lim _{u\to \infty }\left(1+{\frac {1}{u}}\right)^{ku}=\left(\lim _{u\to \infty }\left(1+{\frac {1}{u}}\right)^{u}\right)^{k}=e^{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03bd2c2606b044ae162ad3200bc2f85f35968d8a)
![{\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}=\left[{\begin{matrix}u=e^{x}-1\\x=\ln(1+u)\\x\to 0\\u\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {u}{\ln(1+u)}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44b5743332fc468418ff1a2a73f8a861c50f34c6)
![{\displaystyle \lim _{x\to 0}{\frac {a^{x}-1}{x\ln a}}=\lim _{x\to 0}{\frac {e^{\ln(a^{x})}-1}{x\ln a}}=\lim _{x\to 0}{\frac {e^{x\ln a}-1}{x\ln a}}=\left[{\begin{matrix}u=x\ln a\\u\to 0\\x\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {e^{u}-1}{u}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6b736c2d1000d29dc0f1133acf22341d9509cf5)
![{\displaystyle =\lim _{x\to 0}{\frac {e^{\alpha \ln(1+x)}-1}{\alpha \ln(1+x)}}\cdot 1=\left[{\begin{matrix}u=\alpha \ln(1+x)\\x\to 0\\u\to 0\end{matrix}}\right]=\lim _{u\to 0}{\frac {e^{u}-1}{u}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69c66f44a440fbd5eb2c9a64d1b1a280cec982d2)
