极限
limn→∞an {\displaystyle \lim_{n \rightarrow \infty }{a_n} } n→∞liman = 0
极限的严格定义
称极限 limx→cf(x) {\displaystyle \lim_{x \rightarrow c}{f(x)} } x→climf(x) = L 表示
∀ϵ>0,则比可找到对应的δ>0,0<∣x−c∣<δ⇒∣f(x)−L∣<ϵ \forall ϵ > 0,则比可找到对应的 δ > 0, 0<|x-c| < δ \Rightarrow |f(x) - L| < ϵ ∀ϵ>0,则必须可找到对应的δ>0,0<∣x−c∣<δ⇒∣f(x) -L∣<ϵ
∀ϵ>0∃δ>0,0<∣x−c∣<δ⇒∣f(x)−L∣<ϵ \forall ϵ > 0 \exists δ > 0, 0 < |x-c| < δ \Rightarrow | f(x) - L| < ϵ ∀ϵ>0 ∃δ>0,0<∣x−c∣<δ⇒∣f(x)−L∣<ϵ
极限的求法
直接代入法
limx→2x−4x2−2 {\displaystyle \lim_{x \rightarrow 2}{x - 4 \over x^2 - 2 } } x→2limx2−2x−4
-1
对消法
limx→3x2−9x−3 {\displaystyle \lim_{x \rightarrow 3}{x^2 -9\over x - 3} } x→3limx−3x2−9
limx→2(x−3)(x+3)x−3 {\displaystyle \lim_{x \rightarrow 2}{(x - 3)(x + 3)\over x - 3} } x→2limx−3(x−3)(x+3)
limx→3x+3 {\displaystyle \lim_{x \rightarrow 3}{x + 3} } x→3limx+3
limx→9x−9x−3 {\displaystyle \lim_{x \rightarrow 9}{x - 9 \over \sqrt{x} - 3} } x→9limx−3x−9
limx→9(x−3)(x+3)x−3 {\displaystyle \lim_{x \rightarrow 9}{ (\sqrt{x} - 3)(\sqrt{x} + 3)\over \sqrt{x} - 3} } x→9limx−3(x−3)(x+3)
limx→9(x+3) {\displaystyle \lim_{x \rightarrow 9}{(\sqrt{x} + 3)} } x→9lim(x+3)
当x 趋近无限时
limx→∞4x2+6x−93x2−5x−3 {\displaystyle \lim_{x \rightarrow \infty }{4x^2 + 6x -9 \over 3x^2 -5x -3} } x→∞lim3x2−5x−34x2+6x−9
434 \over 334
limx→∞8x2+7x+83x5−5x−3 {\displaystyle \lim_{x \rightarrow \infty }{8x^2 + 7x + 8 \over 3x^5 -5x - 3} } x→∞lim3x5−5x−38x2+7x+8
0
limx→∞3x4+2x+48x2+5x−3 {\displaystyle \lim_{x \rightarrow \infty }{3x^4 + 2x + 4 \over 8x^2 + 5x -3} } x→∞lim8x2+5x−33x4+2x+4
∞ \infty ∞
xx>x!>ax>xn>logx>sinxx^x > x! > a^x > x^n > \log x > \sin xxx>x!>ax>xn>logx>sinx
limx→∞x100+82x {\displaystyle \lim_{x \rightarrow \infty }{x^{100} + 8 \over 2^x } } x→∞lim2xx100+8
limx→∞logx3x−3 {\displaystyle \lim_{x \rightarrow \infty }{\log x \over 3x - 3 } } x→∞lim3x−3logx
相减式
求极限 limx→∞x2+2x−x {\displaystyle \lim_{x \rightarrow \infty }{\sqrt {x^2 + 2x} - x }} x→∞limx2+2x−x
limx→∞(x2+2x−x)(x2+2x+x)1∗(x2+2x+x) {\displaystyle \lim_{x \rightarrow \infty }{(\sqrt {x^2 + 2x} - x)(\sqrt {x^2 + 2x} + x) \over 1*(\sqrt {x^2 + 2x} + x) }} x→∞lim1∗(x2+2x+x)(x2+2x−x)(x2+2x+x)
limx→∞(x2+2x)−x2x2+2x+x {\displaystyle \lim_{x \rightarrow \infty }{(x^2 + 2x) - x^2 \over \sqrt {x^2 + 2x} + x} } x→∞limx2+2x+x(x2+2x)−x2
limx→∞2xx2+2x+x {\displaystyle \lim_{x \rightarrow \infty }{2x \over \sqrt {x^2 + 2x} + x} } x→∞limx2+2x+x2x
1
夹挤原理
若 h(x)≤f(x)≤g(x)h(x) \leq f(x) \leq g(x)h(x)≤f(x)≤g(x) 且 limx→ah(x)=L,limx→ag(x)=L {\displaystyle \lim_{x \rightarrow a}{h(x)} } = L, {\displaystyle \lim_{x \rightarrow a}{g(x)} } = L x→alimh(x)=L,x→alimg(x)=L 则 limx→af(x)=L {\displaystyle \lim_{x \rightarrow a}{f(x)} } = Lx→alimf(x)=L
求极限 limx→∞[x]x {\displaystyle \lim_{x \rightarrow \infty }{[x] \over x} } x→∞limx[x]
x−1≤[x]≤xx-1 \leq [x] \leq xx−1≤[x]≤x
x−1x≤[x]x≤1{x-1 \over x} \leq {[x] \over x} \leq {1}xx−1≤x[x]≤1 (x > 0)
单边极限
limx→x0+f(x) {\displaystyle \lim_{x \rightarrow x_0^+}{f(x)} } x→x0+limf(x)
右极限
limx→x0−f(x) {\displaystyle \lim_{x \rightarrow x_0^-}{f(x)} } x→x0−limf(x)
左极限
极限若存在，必唯一
若 f(x)=∣x∣xf(x)={|x| \over x }f(x)=x∣x∣ 求极限 limx→0f(x) {\displaystyle \lim_{x \rightarrow 0}{f(x)} } x→0limf(x)
limx→0+∣x∣x=1 {\displaystyle \lim_{x \rightarrow 0^+}{|x| \over x} } = 1x→0+limx∣x∣=1
limx→0−∣x∣x=limx→0−xx=−1 {\displaystyle \lim_{x \rightarrow 0^-}{|x| \over x} } = {\displaystyle \lim_{x \rightarrow 0}{-x \over x} } = -1 x→0−limx∣x∣=x→0limx−x=−1
若 x={3x+6x≤2kx2x>2x = \begin{cases} 3x + 6 &\text x \leq 2 \\ kx^2 &\text x > 2 \end{cases}x={3x+6kx2x≤2x>2 若极限
limx→2−f(x)=12 {\displaystyle \lim_{x \rightarrow 2^-}{f(x)} } = 12 x→2−limf(x)=12
limx→2+f(x)=k∗4 {\displaystyle \lim_{x \rightarrow 2^+}{f(x)} } = k*4x→2+limf(x)=k∗4
k = 3