函数极限计算

函数极限

注意

$ \lim \limits_{x \to x_{0} } {f(x)} = A, x \to x_{0}, x取不到x_{0} $。
以下情况一般要讨论左右两侧极限:
- $ f(x) $在$ x_{0} $两侧的函数表达式不同(分段函数)。
- $ f(x) $的表达式中含有$ e^{\frac{1}{x - a} } $, 且计算$ \lim \limits_{x \to a} {f(x)} $。
- $ f(x) $的表达式中含有$ \arctan{x} 或 \arctan{\frac{1}{x} }$, 且计算$ \lim \limits_{x \to \infty} {f(x)} 或 \lim \limits_{x \to 0} {f(x)}$。

无穷小和无穷大

在 $ \lim{\alpha} = \lim{\beta} = 0 $ ,

若 $ \lim{\frac{\alpha}{\beta} } = 0 $ , 则称 $ \alpha $ 是 $ \beta $ 的高阶无穷小，记为 $ \alpha = o( \beta ) $。

若 $ \lim{\frac{\alpha}{\beta} } = k (k \neq 0, \infty) $ , 则称 $ \alpha $ 是 $ \beta $ 的同阶无穷小，记为 $ \alpha = O( \beta ) $。

若 $ \lim{\frac{\alpha}{\beta} } = 1 $ , 则称 $ \alpha $ 是 $ \beta $ 的等价无穷小，记为 $ \alpha \sim \beta $。

若 $ \lim{\frac{\alpha}{\beta^{k} } } = C (C \neq 0, \infty) $ , 则称 $ \alpha $ 是 $ \beta $ 的k阶无穷小。

注意

并非任何两个无穷小都可以比阶，毕竟数字0不能做分母

极限的四则运算法则

注意

只有在拆开以后的极限都存在时，才能拆开计算。
存在+存在=存在；存在+不存在=不存在；不存在+不存在=不确定。

这里的“存在”，指的是广义上的存在，并不单指“极限存在”，还可以表示“连续性”、“可导性”、“可积性”、“反常积分敛散性”、“无穷级数敛散性”等等。
比如：可导+可导=可导、可导+不可导=不可导、不可导+不可导=不确定、收敛+发散=发散、不可积+不可积=不确定，等等。
事实上，在计算极限 $ \lim {[f(x) \pm g(x)]} $ 时，只要 $ \lim {f(x)} $ 和 $ \lim {g(x)} $ 中至少有一个存在，就能拆开，变成$ \lim {[f(x) \pm g(x)]} = \lim {f(x)} \pm \lim {g(x)} $。
事实上，在计算极限 $ \lim {[f(x)g(x)]} $ 时，若 $ \lim {f(x)} $ 的结果为非零常数，就能先将代入（无论 $ \lim {g(x)} $ 是否存在）。

两个重要极限

$ \lim \limits_{x \to 0} {\frac{\sin{x} }{x} } = 1 $

$ \lim \limits_{x \to \infty} {(1 + x)^{\frac{1}{x} } } = e $

求函数极限的三大方法

等价无穷小

$ x \to 0, x \sim \sin{x} $

$$
\begin{align}
&\because \sin{x} \lt x \lt \tan{x} \\
&\therefore 1 \lt \frac{x}{\sin{x} } \lt \frac{1}{\cos{x} } \\
&\because \lim_\limits{x \to 0}{1} = 1 \\
又&\because \lim_\limits{x \to 0}{\frac{1}{\cos{x} } } = 1 \\
又&\because \lim_\limits{x \to 0}{1} \lt \lim_\limits{x \to 0}{\frac{x}{\sin{x} } } \lt \lim_\limits{x \to 0}{\frac{1}{\cos{x} } } \\
&\therefore \lim_\limits{x \to 0}{\frac{x}{\sin{x} } } = 1
\end{align}
$$

$ x \to 0, x \sim \tan{x} $

$$
\begin{align}
\lim_\limits{x \to 0}{\frac{x}{\tan{x} } } = \lim_\limits{x \to 0}{\frac{x}{\sin{x} }{\cos{x} } } = \lim_\limits{x \to 0}{\frac{x}{\sin{x} } }\lim_\limits{x \to 0}{\cos{x} } = 1
\end{align}
$$

$ x \to 0, x \sim \arcsin{x} $

$$
\begin{align}
&令 t = \arcsin{x}, 则 x = \sin{t} \\
&\lim_\limits{x \to 0}{\frac{x}{\arcsin{x} } } = \lim_\limits{t \to 0}{\frac{\sin{t} }{t} } = \lim_\limits{t \to 0}{ {\frac{1}{\frac{t}{\sin{t} } } } } = \frac{\lim_\limits{t \to 0}{1} }{\lim_\limits{t \to 0}{\frac{t}{\sin{t} } } } = 1
\end{align}
$$

$ x \to 0, x \sim \arctan{x} $

$$
\begin{align}
&令 t = \arctan{x}, 则 x = \tan{t} \\
&\lim_\limits{x \to 0}{\frac{x}{\arctan{x} } } = \lim_\limits{t \to 0}{\frac{\tan{t} }{t} } = \lim_\limits{t \to 0}{ {\frac{\sin{t} }{t\cos{t} } } } = \lim_\limits{t \to 0}{\frac{1}{\frac{t}{\sin{t} }\cos{t} } } = \frac{\lim_\limits{t \to 0}{1} }{\lim_\limits{t \to 0}{\frac{t}{\sin{t} }\lim_\limits{t \to 0}{\cos{t} } } } = 1
\end{align}
$$

$ x \to 0, x \sim \ln{(1 + x)} $

$$
\begin{align}
&令 t = \frac{1}{x}, 则 x = \frac{1}{t} \\
&\lim_\limits{x \to 0}{\frac{x}{\ln{(1 + x)} } } = \lim_\limits{t \to 0}{\frac{\frac{1}{t} }{\ln{(1 + \frac{1}{t})} } } = \lim_\limits{t \to 0}{\frac{1}{t\ln{(1 + \frac{1}{t})} } } = \lim_\limits{t \to 0}{\frac{1}{\ln{(1 + \frac{1}{t})^{t} } } } = 1
\end{align}
$$

$ x \to 0, x \sim e^x - 1 $

$$
\begin{align}
&令t = e^x - 1, 则 x = \ln{(t + 1)} \\
&\lim_\limits{x \to 0}{\frac{x}{e ^ x-1} } = \lim_\limits{t \to 0}{\frac{\ln{(t + 1)} }{t} } = \lim_\limits{t \to 0}{\frac{1}{\frac{t}{\ln{(t + 1)} } } } = 1
\end{align}
$$

$ x \to 0, x \sim \ln{(x + \sqrt{1 + x ^ 2})} $

$$
\begin{align}
&\lim_\limits{x \to 0}{\frac{x}{\ln{(x + \sqrt{1 + x ^ 2})} } } \\
=\quad &\lim_\limits{x \to 0}{\frac{x}{\ln{[1 + (x + \sqrt{1 + x ^ 2} - 1)}]} } \\
=\quad &\lim_\limits{x \to 0}{\frac{x}{x + \sqrt{1 + x ^ 2} - 1} } \\
=\quad &\lim_\limits{x \to 0}{\frac{x[(x - 1) - \sqrt{1 + x ^ 2}]}{[(x - 1) + \sqrt{1 + x ^ 2}][(x - 1) + \sqrt{1 + x ^ 2}]} } \\
=\quad &\lim_\limits{x \to 0}{\frac{x[(x - 1) - \sqrt{1 + x ^ 2}]}{-2x} } \\
=\quad &\lim_\limits{x \to 0}{\frac{(x - 1) - \sqrt{1 + x ^ 2} }{-2} } \\
=\quad &1
\end{align}
$$

$ x \to 0, \frac{1}{2}x^2 \sim 1 - cos{x} $

$$
\lim_\limits{x \to 0}{\frac{\frac{1}{2}x^2}{1- cosx} } = \lim_\limits{x \to 0}{\frac{\frac{1}{2}x^2}{2\sin^2{\frac{x}{2} } } } = \lim_\limits{x \to 0}{\frac{x^2}{4(\frac{x}{2})^2} } = 1
$$

$ x \to 0, (1 + x) ^ {\alpha} - 1 \sim \alpha{x} $

$$
\lim \limits _{x \to 0} { \frac{(1 + x) ^ {\alpha} - 1}{\alpha{x} } } = \lim \limits _{x \to 0} { \frac{e^{\alpha\ln{(1 + x)} } - 1}{\alpha{x} } } = \lim \limits _{x \to 0} { \frac{e^{\alpha\ln{(1 + x)} } - 1}{\alpha{x} } } = \lim \limits _{x \to 0} { \frac{\alpha\ln{(1 + x)} }{\alpha{x} } } = \lim \limits _{x \to 0} { \frac{\alpha{x} }{\alpha{x} } } = 1
$$

等价无穷小二级结论

$ x \to 1, x - 1 \sim \ln{x} $

$$
\lim_\limits{x \to 0}{\frac{x}{\ln{(1 + x)} } } = \lim_\limits{x \to 1}{\frac{x - 1}{\ln{[1 + (x - 1)]} } } = \lim_\limits{x \to 1}{\frac{x - 1}{\ln{x} } } = 1
$$

$ x \to 0, \frac{k}{2}x^{2} \sim 1 - cos^{k}{x} $

$$
\lim\limits_{x \to 0}{\frac{\frac{k}{2}x^2}{1 - cos^k{x} } } = \lim\limits_{x \to 0}{\frac{\frac{k}{2}x^2}{-(cos^k{x} - 1)} } = \lim\limits_{x \to 0}{\frac{\frac{k}{2}x^2}{-{[1 + (cos{x} - 1)]^k - 1} } } = \lim\limits_{x \to 0}{\frac{\frac{k}{2}x^2}{-k(cos{x} - 1)} } = \lim\limits_{x \to 0}{\frac{\frac{k}{2}x^2}{-k(-\frac{1}{2}x^2)} } = 1
$$

$ f(x) \to 0, f(x)g(x) \to 0, [1 + f(x)] ^{g(x)} -1 \sim f(x)g(x) $

$$
\begin{align}
&{\lim \limits _{f(x) \to 0} } \limits _{f(x)g(x) \to 0} {\frac{[1 + f(x)] ^{g(x)} - 1}{f(x)g(x)} } \\
=\quad &{\lim \limits _{f(x) \to 0} } \limits _{f(x)g(x) \to 0} {\frac{e^{g(x)\ln{[1 + f(x)]} } - 1}{f(x)g(x)} } \\
=\quad &{\lim \limits _{f(x) \to 0} } \limits _{f(x)g(x) \to 0} {\frac{ {g(x)\ln{[1 + f(x)]} } }{f(x)g(x)} } \\
=\quad &{\lim \limits _{f(x) \to 0} } \limits _{f(x)g(x) \to 0} {\frac{ {\ln{[1 + f(x)]} } }{f(x)} } \\
=\quad &{\lim \limits _{f(x) \to 0} } \limits _{f(x)g(x) \to 0} {\frac{ {\ln{[f(x)]} } }{f(x)} } \\
=\quad &1
\end{align}
$$

$ f(x) \to 1, g(x) \to \infty, \lim f(x)^{g(x)} = e^{\lim g(x)[f(x) - 1]} $

$$
\begin{align}
&{\lim \limits _{f(x) \to 1} } \limits _{g(x) \to \infty} {f(x)^{g(x)} } \\
=\quad &e ^ { {\lim \limits _{f(x) \to 1} } \limits _{g(x) \to \infty} { {g(x) \ln{f(x)} } } } \\
=\quad &e ^ { {\lim \limits _{f(x) \to 1} } \limits _{g(x) \to \infty} {g(x) [f(x) - 1] } }
\end{align}
$$

由泰勒展开得到的等价无穷小二级结论$ (x \to 0) $

$ x - sinx \sim \frac{1}{6}x^3 $

$$
\begin{align}
\sin{x} &= x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + … + (-1)^n\frac{1}{(2n+1)!}x^{2n + 1} + o(x^{2n+1}) \\
x - \sin{x} &= \frac{1}{3!}x^3 - \frac{1}{5!}x^5 + \frac{1}{7!}x^7 + … + [- (-1)^n\frac{1}{(2n+1)!}x^{2n + 1}] + [-o(x^{2n+1})]
\end{align}
$$

$ x - \ln{(1 + x)} \sim \frac12 x^2 $

$$
\begin{align}
\ln{(1 + x)} &= x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + … + (-1)^n\frac{1}{n + 1}x^{n + 1} + o(x^{n + 1}) \\
x - \ln{(1 + x)} &= \frac{1}{2}x^2 - \frac{1}{3}x^3 + … + [-(-1)^n\frac{1}{n + 1}x^{n + 1}] + [-o(x^{n + 1})]
\end{align}
$$

$ \tan{x} - \sin{x} \sim \frac12 x^3 $

$$
\begin{align}
\tan{x} &= x + \frac{1}{3}x^3 + o(x^3) \\
\sin{x} &= x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + … + (-1)^n\frac{1}{(2n+1)!}x^{2n + 1} + o(x^{2n+1}) \\
\tan{x} - \sin{x} &= \frac12 x^3 + o(x^3) \\ \\
或 \\ \\
\tan{x} - \sin{x} &= \tan{x}(1 - \cos{x}) \sim \frac{1}{2}x^{3}
\end{align}
$$

$ \tan{x} - \arcsin{x} \sim \frac16 x^3 $

$$
\begin{align}
\tan{x} &= x + \frac{1}{3}x^3 + o(x^3) \\
\arcsin{x} &= x + \frac{1}{6}x^3 + o(x^3) \\
\tan{x} - \arcsin{x} &= \frac16 x^3 + o(x^3)
\end{align}
$$

等价推论

$ a \sim b \to 0^{+}, \ln{a} \sim \ln{b} \to -\infty $

$$
\lim \limits_{a \sim b \to 0^{+} } {\frac{\ln{a} }{\ln{b} } } = \lim \limits_{a \sim b \to 0^{+} } {\frac{\ln{(\frac{a}{b} b)} }{\ln{b} } } = \lim \limits_{a \sim b \to 0^{+} } {\frac{\ln{(\frac{a}{b})} }{\ln{b} } } +1 = 1
$$

$ \lim \limits_{x \to 0^{+} } {x \ln{x} } = \lim \limits_{x \to 0^{+} } {x \ln{\sin{x} } } = \lim \limits_{x \to 0^{+} } {x \ln{\arcsin{x} } } = \lim \limits_{x \to 0^{+} } {x \ln{\tan{x} } } = 0 $

$$
\lim \limits_{x \to 0^{+} } {x \ln{x} } = \lim \limits_{x \to 0^{+} } {\frac{\ln{x} }{\frac{1}{x} } } = \lim \limits_{x \to 0^{+} } {\frac{\frac{1}{x} }{-\frac{1}{x^{2} } } } = \lim \limits_{x \to 0^{+} } {-x} = 0 \\
$$

$$
\begin{align}
&\because x \to 0^{+} \\
&\therefore x \sim \sin{x} \sim \arcsin{x} \sim \tan{x} \to 0^{+} \\
&\therefore \ln{x} \sim \ln{\sin{x} } \sim \ln{\arcsin{x} } \sim \ln{\tan{x} } \to -\infty \\
&\therefore \lim \limits_{x \to 0^{+} } {x \ln{x} } = \lim \limits_{x \to 0^{+} } {x \ln{\sin{x} } } = \lim \limits_{x \to 0^{+} } {x \ln{\arcsin{x} } } = \lim \limits_{x \to 0^{+} } {x \ln{\tan{x} } } = 0 \\
\end{align}
$$

注意

如果想对某个部分使用等价无穷小，必须保证该部分与其余部分的全体构成乘除关系，一定不能在加减中使用等价无穷小，也不能局部使用等价无穷小。

洛必达

$ \frac00 $ 型的洛必达法则

注意

$ f(x) $ 和 $ g(x) $ 在 $ x = x_{0} $ 的去心邻域内可导，且分母 $ g(x) \neq 0 $

$ \frac{*}{\infty} $ 型的洛必达法则

注意

洛必达法则中的A，可以是具体的数字，也可以是 $ \infty $ ，结论均成立。
若 $ \lim {\frac{f’(x)}{g’(x)} } $ 是震荡形式的不存在，则 $ \lim {\frac{f(x)}{g(x)} } $ 本身可能存在，也可能不存在，此时应属于“洛必达法则失效”的情况，必须更换计算方法。

泰勒展开

泰勒公式

$$
\begin{align}
f(x) \approx f(x_0) + f’(x_0)(x - x_0) \\
\end{align}
$$

$$
\begin{align}
&f(x) = a_0 + a_1(x-x0) + a2(x - x_0)^2 +…+a_n(x - x_0)^n + R_n(x) \\
&f(x_0) = a_0 \\\\
&f’(x) = a_1 + 2a_2(x - x_0)^1 +…+na_n(x - x_0)^{n - 1} + R_n(x) \\
&f’(x_0) = a_1$ \\\\
&f’’(x) = 2! \times a_2 +…+ n \times (n-1) \times a_n(x - x_0)^{n - 2} + R_n(x) \\
&f’’(x_0) = 2! \times a_2 \\\\
&… \\\\
&f(x) = f(x_0) + f’(x_0)(x - x_0) + \frac{f’’(x_0)}{2!}(x - x_0)^2 + … + \frac{f^{(n)}(x)}{n!}(x - x_0)^n + R_n(x) \\
&R_n(x) = o((x - x_0)^ n) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - x_0)^{n + 1} (\xi \in (x0, x))
\end{align}
$$

麦克劳林公式

$$
令x_0 = 0,f(x) = f(x_0) + f’(x_0)x + \frac{f’’(x_0)}{2!}x^2 + … + \frac{f^{(n)}(x)}{n!}x^n + R_n(x) \\
R_n(x) = \frac{f^{(n + 1)}(\theta{x})}{(n + 1)!}x^{n + 1}(\theta \in (0, 1))
$$

$ e^x = \sum_\limits{n = 0}^{\infty}{\frac{1}{n!}x^n} = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + … + \frac{1}{n!}x^n + o(x^n)$

$$
\begin{align}
&令f(x) = e^x \\ \\
&f(0) = e^x = 1 \\
&f’(0) = e^x = 1 \\
&f’’(0) = e^x = 1 \\
&f’’’(0) = e^x = 1 \\
&e^x = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + … + \frac{a_n}{n!}x^n + o(x^n)
\end{align}
$$

$ a^{x} = \sum_\limits{n = 0}^{\infty}{\frac{\ln^{n}{a} }{n!}x^n} = 1 + x \ln{a} + \frac{\ln^{2}{a} }{2!}x^2 + \frac{\ln^{3}{a} }{3!}x^3 + … + \frac{\ln^{n}{a} }{n!}x^n + o(x^n)$

$$
\begin{align}
&a^{x} \\
= &e^{x \ln{a} } \\
= &1 + x \ln{a} + \frac{\ln^{2}{a} }{2!}x^2 + \frac{\ln^{3}{a} }{3!}x^3 + … + \frac{\ln^{n}{a} }{n!}x^n + o(x^n)
\end{align}
$$

$\sin{x} = \sum_\limits{n = 0}^{\infty}{(-1)^n\frac{1}{(2n+1)!}x^{2n + 1} } = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + … + (-1)^n\frac{1}{(2n+1)!}x^{2n + 1} + o(x^{2n+1})$

$令f(x) = \sin{x}$

$f(0) = \sin{x} = 0$

$f’(0) = \cos{x} = 1$

$f’’(0) = -\sin{x} = 0$

$f’’’(0) = -\cos{x} = -1$

$ f^{(4)}(0) = \sin{x} = 0$

$\sin{x} = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + … + (-1)^n\frac{1}{(2n+1)!}x^{2n + 1} + o(x^{2n+1})$

$\cos{x} = \sum_\limits{n = 0}^{\infty}{(-1)^n\frac{1}{(2n)!}x^{2n} } = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + … + (-1)^n \frac{1}{(2n)!}x^{2n} + o(x^{2n})$

$ \cos{x} = (\sin{x})’ $

$ \cos{x} = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + … + (-1)^n \frac{1}{(2n)!}x^{2n} + o(x^{2n})$

$ \frac{1}{1 - x} = \sum_\limits{n = 0}^{\infty}{x^n} = 1 + x + x^2 + x^3 + … + x^n + o(x^n)$

$令 f(x) = \frac{1}{1 - x} $

$ f(0) = \frac{1}{1 - x} = 1 $

$ f’(0) = \frac{1}{(1 - x) ^ 2} = 1$

$ f’’(0) = \frac{2!}{(1 - x) ^ 3} = 2$

$ f’’’(0) = \frac{3!}{(1 - x) ^ 3} = 3$

$ \frac{1}{1 - x} = 1 + x + x^2 + x^3 + … + x^n + o(x^n)$

$ \frac{1}{1 + x} = \sum_\limits{n = 0}^{\infty}{(-1)^nx^n} = 1 - x + x^2 - x^3 + … + (-1)^nx^n + o(x^n)$

$ \frac{1}{1 + x} = \frac{1}{1 - (-x)}$

$ \frac{1}{1 + x} = 1 - x + x^2 - x^3 + … + (-1)^nx^n + o(x^n)$

$ \ln{(1 + x)} = \sum \limits _{n = 0} ^{\infty}(-1)^n\frac{1}{n + 1}x^{n + 1} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + … + (-1)^n\frac{1}{n + 1}x^{n + 1} + o(x^{n + 1})$

$ [\ln{(1 + x)}]’ = \frac{1}{1 + x} $

$ \ln{(1 + x)} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + … + (-1)^n\frac{1}{n + 1}x^{n + 1} + o(x^{n + 1})$

$ (1 + x) ^ {\alpha} = \sum\limits_{n = 0}^{\infty}{C_{\alpha}^{n}x^n} = C_{\alpha}^{0} x^0 + C_{\alpha}^{1} x^1 + C_{\alpha}^{2} x^2 + … + C_{\alpha}^{n} x^{n} + o(x^{n})$

$ (1 + x)^{\alpha} = C_{\alpha}^{0} 1^{\alpha}x^0 + C_{\alpha}^{1} 1^{\alpha - 1}x^1 + C_{\alpha}^{2} 1^{\alpha - 2}x^2 + … + C_{\alpha}^{n} 1^{\alpha - n}x^{n} + o(x^{n}) \\ = C_{\alpha}^{0} x^0 + C_{\alpha}^{1} x^1 + C_{\alpha}^{2} x^2 + … + C_{\alpha}^{n} x^{n} + o(x^{n})$

$ (1 + x)^{\frac{1}{x} } = e ( 1 - \frac{1}{2} x + \frac{11}{24}x^{2} - \frac{7}{16} x^{3} ) + o(x^{3})$

$$
\begin{align}
&(1 + x)^{\frac{1}{x} } \\
=\quad &e ^ {\frac{1}{x} \ln{(1 + x)} } \\
=\quad &e ^ {\frac{1}{x} [x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + ··· + (-1)^{n + 1} \frac{1}{n} x^{n}]} \\
=\quad &e ^ {1 - \frac{1}{2} x + \frac{1}{3} x^{2} - \frac{1}{4} x^{3} + ··· + (-1)^{n + 1} \frac{1}{n} x^{n - 1} } \\
=\quad &e · e^{- \frac{1}{2} x} · e^{\frac{1}{3} x^{2} } · e^{- \frac{1}{4} x^{3} } ····· e^{(-1)^{n + 1} \frac{1}{n} x^{n - 1} } \\
=\quad &e [1 - \frac{1}{2} x + \frac{1}{2!}(- \frac{1}{2} x)^{2} + ··· + \frac{1}{n!}(- \frac{1}{2} x)^{n} ] [1 + \frac{1}{3} x^{2} + \frac{1}{2!}(\frac{1}{3} x^{2})^{2} + ··· + \frac{1}{n!}(\frac{1}{3} x^{2})^{n}] \\
&[1 - \frac{1}{4} x^{3} + \frac{1}{2!}(- \frac{1}{4} x^{3})^{2} + ··· + \frac{1}{n!}(- \frac{1}{4} x^3)^{n} ] ····· \\
&\{ 1 + (-1)^{n + 1} \frac{1}{n} x^{n - 1} + \frac{1}{2!}[(-1)^{n + 1} \frac{1}{n} x^{n - 1}]^{2} + ··· + \frac{1}{n!}[(-1)^{n + 1} \frac{1}{n} x^{n - 1}]^{n} \} \\
=\quad &e \{ 1 - \frac{1}{2} x + [\frac{1}{3} + \frac{1}{2!}· (- \frac{1}{2})^{2}] x^{2} + [-\frac{1}{4} + \frac{1}{3} · (- \frac{1}{2}) + \frac{1}{3!}·(-\frac{1}{2})^{3}] x^{3} + o(x^{3}) \} \\
=\quad &e ( 1 - \frac{1}{2} x + \frac{11}{24}x^{2} - \frac{7}{16} x^{3} ) + o(x^{3})
\end{align}
$$

$ \arctan{x} = \sum \limits _{n = 0} ^{\infty} (-1)^n \frac{1}{2n + 1} x^{2n + 1} = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 + … + (-1)^n \frac{1}{2n + 1} x^{2n + 1} + o(x^{2n + 1})$

$ \frac{1}{1 + x ^ 2} = \frac{1}{1 + (x ^ 2)}$

$ \frac{1}{1 + x ^ 2} = 1 - x^2 + x^4 - x^6 + … + (-1)^nx^{2n} + o(x^{2n}) $

$ (\arctan{x})’ = \frac{1}{1 + x ^ 2}$

$ \arctan{x} = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 + … + (-1)^n \frac{1}{2n + 1} x^{2n + 1} + o(x^{2n + 1}) $

$ \tan{x} = \sum\limits_{n = 0}^{\infty}\frac{T_n}{(2n + 1)!}x^{2n+1}$ $(T_n 为正切系数)$ $ = x + \frac{1}{3}x^3 + o(x^3)$

$ 令f(x) = \tan{x} $

$ f(0) = \tan{x} = 0 $

$ f’(0) = \sec^2{x} = 1 $

$ f’’(0) = 2\sec^2{x}\tan{x} = 0 $

$ f’’’(0) = 2\sec^4{x} + 4 \sec^2{x}\tan^2{x} = 2 $

$ \tan{x} = x + \frac{2}{3!}x^3 + \frac{16}{5!}x^5 + \frac{272}{7!}x^7 + … + \frac{T_n}{(2n + 1)!}x^{2n+1} + o(x^{2n + 1}) $ $ (T_n 为正切系数)$

$ \arcsin{x} = \sum\limits_{n = 0}^{\infty}\frac{T_n}{(2n + 1)!}x^{2n+1} $ $ = x + \frac{1}{6}x^3 + o(x^3) $ $ (T_n = \left\{\begin{aligned} 1 \quad n=0 \\ \prod_{n = 0}^{n - 1}{(2n+1)^2} \quad n > 0 \end{aligned} \right.)$

$$
\begin{align}
&令f(x) = \arcsin{x} \\
&f(0) = \arcsin{x} = 0 \\
&f’(0) = \frac{1}{\sqrt{1 - x^2} } = 1 \\
&f’’(0) = \frac{x}{(1 - x^2)^{\frac32} } = 0 \\
&f’’’(0) = \frac{3x^2}{(1 - x^2)^{\frac52} } + \frac{1}{(1 - x^2)^\frac32} = 1 = 1 \times 1 \\
&f^{(4)}(0) = \frac{9 x}{\left(1-x^2\right)^{5/2} }+\frac{15 x^3}{\left(1-x^2\right)^{7/2} } = 0 \\
&f^{(5)}(0) = \frac{105 x^4}{\left(1-x^2\right)^{9/2} } + \frac{90 x^2}{\left(1-x^2\right)^{7/2} }+\frac{9}{\left(1-x^2\right)^{5/2} } = 9 = 1 \times 1 \times 3 \times 3 \\
&f^{(6)}(0) = \frac{945 x^5}{\left(1-x^2\right)^{11/2} }+\frac{1050 x^3}{\left(1-x^2\right)^{9/2} } + \frac{225 x}{\left(1-x^2\right)^{7/2} } = 0 \\
&f^{(7)}(0) = \frac{10395 x^6}{\left(1-x^2\right)^{13/2} }+\frac{14175 x^4}{\left(1-x^2\right)^{11/2} } + \frac{4725 x^2}{\left(1-x^2\right)^{9/2} }+\frac{225}{\left(1-x^2\right)^{7/2} } = 225 = 1 \times 1 \times 3 \times 3 \times 5 \times 5 \\
&f^{(8)}(0) = \frac{135135 x^7}{\left(1-x^2\right)^{15/2} }+\frac{218295 x^5}{\left(1-x^2\right)^{13/2} }+\frac{99225 x^3}{\left(1-x^2\right)^{11/2} } + \frac{11025 x}{\left(1-x^2\right)^{9/2} } = 0 \\
&f^{(9)}(0) = \frac{2027025 x^8}{\left(1-x^2\right)^{17/2} }+\frac{3783780 x^6}{\left(1-x^2\right)^{15/2} }+\frac{2182950 x^4}{\left(1-x^2\right)^{13/2} } + \frac{396900 x^2}{\left(1-x^2\right)^{11/2} }+\frac{11025}{\left(1-x^2\right)^{9/2} } = 11025 = 1 \times 1 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7 \\
&f^{(10)}(0) = \frac{34459425 x^9}{\left(1-x^2\right)^{19/2} }+\frac{72972900 x^7}{\left(1-x^2\right)^{17/2} }+\frac{51081030 x^5}{\left(1-x^2\right)^{15/2} }+\frac{13097700 x^3}{\left(1-x^2\right)^{13/2} } + \frac{893025 x}{\left(1-x^2\right)^{11/2} } = 0 \\
&f^{(11)}(0) = \frac{654729075 x^{10} }{\left(1-x^2\right)^{21/2} }+\frac{1550674125 x^8}{\left(1-x^2\right)^{19/2} }+\frac{1277025750 x^6}{\left(1-x^2\right)^{17/2} }+\frac{425675250 x^4}{\left(1-x^2\right)^{15/2} } + \frac{49116375 x^2}{\left(1-x^2\right)^{13/2} }+\frac{893025}{\left(1-x^2\right)^{11/2} } = 893025 = 1 \times 1 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7 \times 9 \times 9 \\
&f^{(11)}(0) = \frac{13749310575 x^{11} }{\left(1-x^2\right)^{23/2} }+\frac{36010099125 x^9}{\left(1-x^2\right)^{21/2} }+\frac{34114830750 x^7}{\left(1-x^2\right)^{19/2} }+\frac{14047283250 x^5}{\left(1-x^2\right)^{17/2} }+\frac{2341213875 x^3}{\left(1-x^2\right)^{15/2} } + \frac{108056025 x}{\left(1-x^2\right)^{13/2} } = 0 \\
&f^{(12)}(0) = \frac{316234143225 x^{12} }{\left(1-x^2\right)^{25/2} }+\frac{907454497950 x^{10} }{\left(1-x^2\right)^{23/2} }+\frac{972272676375 x^8}{\left(1-x^2\right)^{21/2} }+ \frac{477607630500 x^6}{\left(1-x^2\right)^{19/2} } + \frac{105354624375 x^4}{\left(1-x^2\right)^{17/2} } + \frac{8428369950 x^2}{\left(1-x^2\right)^{15/2} }+\frac{108056025}{\left(1-x^2\right)^{13/2} } = 108056025 = 1 \times 1 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7 \times 9 \times 9 \times 11 \times 11 \\
&\arcsin{x} = x + \frac{1}{3!}x^3 + \frac{9}{5!}x^5 + \frac{272}{7!}x^7 + … + \frac{T_n}{(2n +
1)!}x^{2n+1} + o(x^{2n + 1}) T_n = \left\{\begin{aligned} 1 \quad n=0 \\ \prod_{n = 0}^{n - 1}{(2n+1)^2} \quad n > 0 \end{aligned} \right. \\
\end{align}
$$

补充

函数极限计算中最基本的技巧与手法

对于零因子可以使用等价无穷小替换，对于非零因子可以先算出来
见到指数函数相减，可以提公因式，构造等价无穷小(1-1型)
见到根号差，就用有理化
见到幂指函数，一般都要取指对数
对于“ $ \infty $ - $ \infty $”的极限，若有分母，可以先通分
利用阶的吸收律，简化运算
一个简单而常用的等价无穷大
洛必达法则可以帮助我们“降阶”
泰勒展开，是求函数极限的杀手锏

注意

当 $ n \to \infty $, $ n^{n} \gg n! \gg e^{n}(指数) \gg n^{2}(幂函数) \gg \ln{n} $

等价无穷小的两种灵活实用

逆用等价无穷小
$$
1 - \cos^{k}{x} \sim -\ln{\cos^{k}{x} } \sim (-k) \ln{\cos{x} } \sim (-k)(\cos{x} - 1) \sim \frac{k}{2}x^2
$$
添项减项

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