数学常用方法

函数极限

基础方法与技巧

经典错解

\begin{aligned}&\int \frac{\frac{1}{x^2} + 1}{x^2 + \frac{1}{x^2} }\, {\rm d}x \Longrightarrow \int \frac{1}{(x - \frac{1}{x} ) ^ 2 + 2}\,{\rm d}(x - \frac{1}{x} ) = \frac{1}{\sqrt 2} \arctan{\frac{x - \frac{1}{x} }{\sqrt 2}} + C\end{aligned}

​ 由于原函数为连续函数，而积分函数在x = 0处左右极限不等

​ 正确解法为

\begin{aligned}&\because\lim_{x \to 0^+}\arctan{\frac{x - \frac{1}{x} }{\sqrt 2}} = \frac{-1}{\sqrt 2}\pi \\& \quad \lim_{x \to 0^-}\arctan{\frac{x - \frac{1}{x} }{\sqrt 2}} = \frac{1}{\sqrt 2}\pi \\&\therefore F(x) = \begin{cases}\arctan{\frac{x - \frac{1}{x} }{\sqrt 2}} + \frac{1}{\sqrt 2}\pi, \quad x > 0\\ 0 \qquad \qquad \qquad\qquad ,\quad x = 0 \\\arctan{\frac{x - \frac{1}{x} }{\sqrt 2}} - \frac{1}{\sqrt 2}\pi, \quad x < 0 \end{cases} \\&\therefore \int \frac{\frac{1}{x^2} + 1}{x^2 + \frac{1}{x^2} }\, {\rm d}x = F(x) + C\end{aligned} 1.添项减项凑出需要的结构

例1.1

\begin{aligned}提示:&A \cdot B - 1 = B(A - 1) + B - 1\end{aligned}

\begin{aligned}&求\lim_{x \to 0} \dfrac{\sqrt[m]{1 + \alpha x } \cdot \sqrt[n]{1 + \beta x } - 1}{x}\end{aligned} 解答

• 解法 1
• 解法 2

\begin{aligned}解法1: \\&原式=\lim_{x \to 0} \dfrac{\sqrt[m]{1 + \alpha x } \cdot \sqrt[n]{1 + \beta x } - \sqrt[n]{1 + \beta x } + \sqrt[n]{1 + \beta x } - 1}{x} \\&=\lim_{x \to 0} \dfrac{ \sqrt[n]{1 + \beta x } \cdot (\sqrt[m]{1 + \alpha x } - 1)}{x} + \lim_{x \to 0} \dfrac{\sqrt[n]{1 + \beta x } - 1}{x} \\& \qquad \qquad \qquad 又 \lim_{x \to 0} \sqrt[n]{x + \beta x} = 1 \\ &=\lim_{x \to 0} \dfrac{\sqrt[m]{1 + \alpha x } - 1}{x} + \lim_{x \to 0} \dfrac{\sqrt[n]{1 + \beta x } - 1}{x}\\ &=\frac{\alpha}{m} + \frac{\beta}{n} \\ \end{aligned}

\begin{aligned}解法2: \\&\because x \to 0, \ln (x + 1) \sim x \Rightarrow \ln x \sim x - 1 \\ &原式=\lim_{x \to 0}\ln {(1 + \alpha x) ^ {\frac{1}{m} } \cdot (1 + \beta x) ^ {\frac{1}{n} } } \\&= \lim_{x \to 0} \frac{1}{m} \cdot \ln (1 + \alpha x) + \frac{1}{n} \cdot \ln (1 + \beta x) \\&=\frac{\alpha}{m} + \frac{\beta}{n} \\\end{aligned}

例1.2

\begin{aligned}&\lim_{x \to 0} \frac{1 - \cos x \cdot \cos 2x ....\cos {\rm n}x}{x ^ 2}\end{aligned} 解答

• 解法 1
• 解法 2

\begin{aligned}解法1: \\&原式=\lim_{x \to 0} \dfrac{1 - \cos x + \cos x - \cos x \cdot \cos 2x ....\cos {\rm n}x}{x ^ 2} \\&= \frac{1}{2} + \lim_{x \to 0}\dfrac{ \cos x (1 - \cos 2x ....\cos {\rm n}x) }{x ^ 2} \\&= \frac{1}{2} + \lim_{x \to 0}\dfrac{ 1 - \cos 2x ....\cos {\rm n}x}{x ^ 2} \\&= \frac{1}{2}(1^2 + 2^2 + 3^2 + .....+ n ^ 2) \\&= \frac{1}{2} \cdot \dfrac{n(n + 1)(n + 2)}{6}\end{aligned}

\begin{aligned}解法2: \\&原式=\lim_{x \to 0} -\dfrac{\cos x \cdot \cos 2x ....\cos {\rm n}x - 1}{x ^ 2} \\&\quad \qquad \because \lim_{x \to 0} \ln \cos x \sim \cos x - 1 \\&= \lim_{x \to 0} \dfrac{\ln \cos x + .... \ln \cos {\rm n} x}{x ^ 2} \\&= \lim_{x \to 0} -\sum_{k = 1} ^ n \frac{\ln \cos kx}{x ^ 2} = \frac{1}{2} \cdot \dfrac{n(n + 1)(n + 2)}{6} \\\end{aligned}

例1.3

\begin{aligned}证明调和级数式发散的: \\&\sum_{n = 1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n} \\\end{aligned}

提示:\lim_{n \to 0} \frac{1}{n} \sim \lim_{n \to 0} \ln (1 + \frac{1}{n} )

解答

\begin{aligned}&\sum_{n = 1}^{\infty} \frac{1}{n} = \sum_{n = 1}^{\infty} \ln (1 + \frac{1}{n}) \\&=\lim_{n \to \infty} \sum_{k = 1} ^ {n} \ln {(1 + \frac{1}{k}) } = \lim_{n \to \infty} \sum_{k = 1} ^ {n} \ln \frac{k + 1}{k} \\ &=\lim_{n \to \infty} (\ln \frac{2}{1} + \ln \frac{3}{2} + \ln \frac{4}{1} + ... \ln \frac{n + 1}{n}) \\ &=\lim_{n \to \infty} \ln (n + 1) = +\infty \\\end{aligned}

例1.4

\begin{aligned}&求极限 \lim_{x \to 0} \dfrac{1 - \cos x \cdot \sqrt{\cos 2x} \cdot \sqrt[3]{\cos 3x} ... \sqrt[n]{\cos nx} }{x ^ 2}\end{aligned} 解答

\begin{aligned}&原式=\lim_{x \to 0} - \dfrac{\ln (\cos x \cdot \sqrt{\cos 2x} \cdot \sqrt[3]{\cos 3x} ... \sqrt[n]{\cos nx}) }{x^2} \\&=\lim_{x \to 0} -\dfrac{\cos {x - 1} + \frac{1}{2}(\cos {2x} - 1) + \frac{1}{3}(\cos {3x} - 1) + ... + \frac{1}{n}(\cos {nx} - 1 ) }{x^2} \\&=\lim_{x \to 0} \dfrac{\frac{1}{2} x ^ 2 + \frac{1}{2} (2x) ^ 2 + \frac{1}{3} (3x) ^ 2 + ... + \frac{1}{n}(nx)^2 }{x^2} \\&=\lim_{x \to 0} \frac{1}{2} + \dfrac{(n + 2)(n - 1)}{2} \\\end{aligned}

例1.5

\begin{aligned}&\lim_{x \to 0} \dfrac{(1 - \cos x)[x - \ln (1 + \tan x)]}{x ^ 4} \\\end{aligned}

\begin{aligned}提示:&\lim_{x \to 0} x - \ln (1 + x) \sim \frac{x ^ 2}{2} \\&\lim_{x \to 0} x - \tan x \sim -\frac{x^3}{3} \\ \end{aligned}

解答

\begin{aligned}&原式 = \lim_{x \to 0} \dfrac{\frac{1}{2} x ^ 2 [x - \ln (1 + \tan x)] }{x ^ 4} \\ &=\frac{1}{2}\lim_{x \to 0} \dfrac{x - \tan x + \tan x - \ln (1 + \tan x) }{x ^ 2} \\&=\frac{1}{2}\lim_{x \to 0} \left[ \frac{x - \tan x}{x ^ 2} + \frac{\tan x - \ln (1 + \tan x)}{x ^ 2} \right] \\&=\frac{1}{2}\lim_{x \to 0} [0 + \frac{\frac{1}{2} \tan x}{x ^ 2}] = \frac{1}{4} \\\end{aligned}

2.凡是幂指函数，大概率就取指对数

例2.1

\begin{aligned}求极限:{\LARGE \lim_{x \to 0}(\frac{\sin x}{x}) ^ {\frac{1}{1 - \cos x} } } \\\end{aligned} 解答

\begin{aligned}&= {\Large {\rm e} ^ {\lim_{x \to 0} \frac{1}{\cos x} \cdot \ln \frac{\sin x}{x} } } \\&= {\Large{\rm e} ^ {\lim_{x \to 0} \frac{1}{\frac{1}{2} x ^ 2} \cdot (\frac{\sin x}{x} - 1)} } \\&= {\Large{\rm e} ^ {2\lim_{x \to 0} \frac{\sin x - x}{x ^ 3} } }\\&= {\Large e ^ {- \frac{1}{3} } }\\\end{aligned}

例2.2

\begin{aligned}求极限: {\Large \lim_{x \to 0} \frac{1}{x^3}[(\frac{2 + \cos x}{3}) ^ x - 1] } \\\end{aligned} 解答

\begin{aligned}原式&{=\Large \lim_{x \to 0} \frac{1}{x ^ 3}[{\rm e} ^ {x \cdot \ln {\frac{2 + \cos x}{3} } } - 1] } \\&{=\Large \lim_{x \to 0} \dfrac{\ln {\frac{2 + \cos x}{3} - 1 } }{x ^ 2} } \\&{=\Large \lim_{x \to 0} \dfrac{\frac{\cos x - 1}{3} }{x ^ 2} = - \frac{1}{6} } \\\end{aligned}

例2.3

\begin{aligned}求极限: {\Large \lim_{x \to \infty}[ \dfrac{(1 + \frac{1}{x}) ^ x}{ {\rm e} } ] ^ x} \\\end{aligned} 解答

\begin{aligned}原式&{\Large =\lim_{x \to \infty} \dfrac{ {\rm e} ^ {x ^ 2 \ln (1 + \frac{1}{x} )} } { {\rm e} ^ x} } \\ &{\Large = \lim_{x \to \infty} {\rm e} ^ {[x ^ 2 \ln (x + \frac{1}{x}) - x ]} } \\对&\ln (1 + \frac{1}{x}) 进行泰勒展开得: \\&{\Large = \lim_{x \to \infty} {\rm e} ^ {[x ^ 2 (\frac{1}{x} - \frac{1}{x ^ 2} \cdot \frac{1}{2} ) - x]} = {\rm e} ^ {- \frac{1}{2} } } \end{aligned}

例2.4

\begin{aligned}求极限: {\Large \lim_{x \to 0} \dfrac{(3 + 2 \tan x) ^ x - 3 ^ x}{3 \sin^2 x + x ^ 3 \cos \frac{1}{x} } } \\\end{aligned}

有时也不一定要取指对数:

吸收率: 低阶 + 高阶 $\sim$ 低阶

\begin{aligned}&a ^ b - c ^ d(难算) \\&a ^ b - a ^ c = a ^ c(a ^ {b - c} - 1) \\&a ^ c - b ^ c = b ^ c( (\frac{a}{b}) ^ c - 1) \\\end{aligned}

解答

泰勒展开求极限

几道经典难题

连续性与间断点

介值定理+零点定理+压缩映像原理

连续函数的"介值定理+零点定理"