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}} Cend{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}&becauselim_{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 qquadqquad ,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) Cend{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 - 1end{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} ^ {2lim_{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"
