常数与无穷小的乘积是无穷小
有限个无穷小的乘积是无穷小
$M = m^2 + n^2 - 4m + 2n = (m^2 - 4m + 4) + (n^2 +2n + 1) - 5 = (m - 2)^2 + (n + 1)^2 - 5$
(1) $lim[f(x)\pm g(x)] = limf(x)\pm limg(x)=A\pm B$
(2) $lim[f(x) . g(x)] = limf(x) * limg(x)=A * B$
(3) $B\neq 0, lim{f(x) \over g(x)} = {limf(x) \over limg(x)}=\frac AB$
$\quad lim[cf(x)] = climf(x)$
$\quad lim[f(x)]^n = [limf(x)]^n$
$lim_{x\to 1}(2x-1) = lim2x-lim1 = 2limx -1 = 2*1-1 = 1$
$lim_{x\to 2}{x^3-1 \over x^2-5x+3} = {lim_{x\to 2}(x^3-1) \over lim_{x\to 2}(x^2-5x+3)} = {[lim_{x\to 2}x]^3 - lim_{x\to 2}1 \over [lim_{x \to 2}x]^2 -5lim_{x\to 2}x +3} = {8 -1 \over 4 - 10 + 3} = -\frac 73 $
$lim_{x\to \infty}{3x^2-2x-1 \over 2x^3-x^2+5} = lim_{x\to \infty}{\frac 3x - \frac2{x^2} - \frac 1{x^3} \over 2 - \frac 1x + \frac 5{x^3}} = \frac 02 = 0 $
$lim_{x\to 0}{sinx \over x} = 1$
$lim_{x\to \infty}(1+ \frac 1x)^x = e$
$lim_{x\to 0}{x \over log_a(1+x)} = ln a$
$lim_{x\to 0}{(1+x)^n -1 \over x} = a$
$ lim_{x\to 0}{a^x-1 \over x} = lna $
复合函数,及分子分母无穷小
$$ lim_{x\to 3}\sqrt{x-3 \over x^2-9} = \sqrt{lim_{x\to 3}{x-3 \over x^2-9}} = \sqrt{lim_{x\to 3}{1 \over 2x}} = \sqrt \frac 16 = {\sqrt 6 \over 6} $$
$$ lim_{x\to 0}{log_a(1+x) \over x} = lim_{x\to 0}log_a(1 + x)^{\frac 1x} = log_alim_{x\to 0}(1 + x)^{\frac 1x} = log_ae = {1 \over lna} $$
求$ lim_{x\to 0}{a^x-1 \over x}$
解：令$a^x-1=t, 则x=log_a(1+t), 当x\to 0, t\to 0, 于是$
$$ lim_{x\to 0}{a^x - 1 \over x} = lim_{t\to 0}{t \over log_a(1+t)} = ln a $$
$$ f'(x_0) = lim_{\Delta x \to 0}{\Delta y \over \Delta x} = lim_{\Delta x \to 0}{f(x_0 + \Delta x) - f(x_0) \over \Delta x } = lim_{h \to 0}{f(x_0 + h) - f(x_0) \over h } $$
$(x^n)' = nx^{n-1}$
(sinx)' = cosx
(cosx)' = -sinx
$(tan x)' = sec^2 x$
$(cot x)' = -csc^2 x$
$(log_ax)' = {1 \over xlna}$
$(lnx)' = \frac 1x$
$(arctan x)' = {1 \over 1+x^2}$
$(arcsin x)' = {1 \over \sqrt{1-x}}$
(secx)' = secxtanx
(cscx)' = -cscxcotx
$(e^x)' = e^x$
$({a^x \over lna})' = a^x $
$f(x_0+\Delta x) \approx f(x_0) + f'(x_0)\Delta x$
$$ sin30^o30' = sin({\pi \over 6} + {\pi \over 360}) \approx sin{\pi \over 6} + cos{\pi \over 6} * {\pi \over 360} = \frac 12 + {\sqrt 3 \over 2} * {\pi \over 360} = 0.5076 $$
$$ \sqrt{1.05} \approx 1 + \frac 12(0.05) = 1.025 $$
$(1+x)^a \approx 1 + ax$
$sinx \approx x$
$tanx \approx x$
$e^x \approx 1 + x$
$ln(1+x) \approx x$
如果函数满足
(1) 在闭区间[a, b]上连续
(2) 在开区间(a, b)上可导
那么在(a, b)内至少有一点$\xi(a < \xi < b)， 使如下等式成立：$
$$ f(b) - f(a) = f'(\xi)(b - a) $$
几何上的意义：曲线上有一点C的切线平行于弦AB
如果函数f(x)和F(x)满足
(1) 在闭区域[a, b]上连续
(2) 在开区间(a, b)内可导
(3) 对任意$x\subseteq(a, b), F(x)\neq 0$
那么在(a, b)内至少有一点$\xi$, 使等式成立
$$ {f(b) - f(a) \over F(b) - F(a)}= {f'(\xi) \over F'(\xi)} $$
特例：若F(x) = x, 则F(b) - F(a) = b - a, F'(x)=1, 得拉格朗日中值公式：
$$ f(b) - f(a) = f'(\xi)(b - a) $$
定理 (同样适用于$x\to \infty$)
(1) 当$x\to a$时函数f(x)及F(x)都趋于0
(2) 在点a的某去心邻域内，f'(x)及F'(x)都存在，且F'(x)$\neq$0
(3) $lim_{x\to a}f'(x) \over lim_{x\to a}F'(x)$存在或无穷大
则
$$ lim_{x\to a}{f(x) \over F(x)} = lim_{x\to a}{f'(x) \over F'(x)} $$
$$ lim_{x\to 0}{sinax \over sin bx} = lim_{x\to 0}{acos ax \over bcos bx} = \frac AB $$
$$ lim_{x\to 1}{x^3-3x+2 \over x^3-x^2-x+1} = lim_{x\to 1}{3x^2-3 \over 3x^2-2x-1} = lim_{x\to 1}{6x \over 6x-2} = \frac 64 = \frac 32 $$
6x, 6x-2不趋近于0，所以不能继续求导
如果函数f(x)在x0的某个邻域U(x0)内具有n+1阶可导
$$ f(x) = f(x_0) + f'(x_0)(x-x_0) + {f''(x_0) \over 2!}(x-x_0)^2 + \cdots + {f^{(n)}(x_0) \over n!}(x - x_0)^n + R_n(x) $$
$\xi$是x0与x间的某个值
$$ R_n(x) = {f^{(n+1)}(x_0) \over (n+1)!}(x - x_0)^{n+1} $$
$$ e^x = 1 + x + {x^2 \over 2!} + \cdots + {x^n \over n!} + {e^{\theta x} \over (n+1)!}x^{n+1} $$
当x = 1时
$$ e \approx 1 + 1 + {1 \over 2!}+ \cdots + {1 \over n!} $$
误差值：
$$ |R_n| < {e \over (n+1)!} < {3 \over (n+1)!} $$
当n = 10时，算出e$\approx$2.718282, 其误差不超过$10^{-6}$
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
\begin{align} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{align}