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 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 − 1 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 − 1 α x = lim x → 0 e α ln ( 1 + x ) − 1 α x = lim x → 0 e α ln ( 1 + x ) − 1 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 ln ( 1 + x ) − 1 α ln ( 1 + x ) ⋅ 1 = [ u = 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.
Из Википедии, бесплатной энциклопедии
![{\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)
