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第42课 — 极限的代数性质

Limite da soma, produto, quociente, potência. Composição. Confronto. Ferramentas operacionais.

Used in: 2.º ano do EM (16-17 anos) · Equiv. Math II japonês (極限の性質) · Equiv. Oberstufe Grenzwertregeln alemão

lim(f±g)=limf±limg,lim(fg)=limflimg,lim(f/g)=limf/limg\lim (f \pm g) = \lim f \pm \lim g, \quad \lim (fg) = \lim f \cdot \lim g, \quad \lim (f/g) = \lim f / \lim g
Choose your door

Rigorous notation, full derivation, hypotheses

运算性质

limxaf(x)=L\lim_{x \to a} f(x) = Llimxag(x)=M\lim_{x \to a} g(x) = M。则:

运算结果
lim(f±g)\lim (f \pm g)L±ML \pm M
lim(cf)\lim (cf)cc 常数cLcL
lim(fg)\lim (fg)LMLM
lim(f/g)\lim (f/g)M0M \neq 0L/ML/M
limfn\lim f^nnNn \in \mathbb{N}LnL^n
limfn\lim \sqrt[n]{f}Ln\sqrt[n]{L}(定义时)
$\limf

复合

limxag(x)=b\lim_{x \to a} g(x) = bffbb 连续,则 limxaf(g(x))=f(b)\lim_{x \to a} f(g(x)) = f(b)

注意:无连续性,复合可能失败。反例:g(x)=0g(x) = 0 常数,ff 在 0 处可去间断。

夹逼定理(三明治)

aa 邻域(除 aa 外)g(x)f(x)h(x)g(x) \leq f(x) \leq h(x)limg=limh=L\lim g = \lim h = L,则 limf=L\lim f = L

经典应用

limx0xsin(1/x)=0\lim_{x \to 0} x \sin(1/x) = 0 由夹逼:xxsin(1/x)x-|x| \leq x \sin(1/x) \leq |x|limx=0\lim |x| = 0

何时不行

  • 0/0、∞/∞:不定式,需操作。
  • 分母 → 0:可能 ±\pm \infty 或不存在。
  • 与不连续 ff 复合:需谨慎。

Exercise list

42 exercises · 10 with worked solution (25%)

Application 20Understanding 6Modeling 6Challenge 4Proof 6
  1. Ex. 42.1Application
    limx2(x2+3x1)\lim_{x \to 2} (x^2 + 3x - 1)。(答:9。)
  2. Ex. 42.2ApplicationAnswer key
    limx1x2+3\lim_{x \to 1} \sqrt{x^2 + 3}
  3. Ex. 42.3ApplicationAnswer key
    limx0(x+1)/(x2+1)\lim_{x \to 0} (x + 1)/(x^2 + 1)
  4. Ex. 42.4ApplicationAnswer key
    limx3(x29)/(x3)\lim_{x \to 3} (x^2 - 9)/(x - 3)。(答:6。)
  5. Ex. 42.5Application
    limx0(1cosx)/x2\lim_{x \to 0} (1 - \cos x)/x^2
  6. Ex. 42.6Application
    limx0sin(3x)/x\lim_{x \to 0} \sin(3x)/x。(答:3。)
  7. Ex. 42.7Application
    limx0tanx/x\lim_{x \to 0} \tan x / x
  8. Ex. 42.8Application
    limx(2x23x+1)/(x2+5)\lim_{x \to \infty} (2x^2 - 3x + 1)/(x^2 + 5)。(答:2。)
  9. Ex. 42.9Application
    limx(x+1)/(x2+1)\lim_{x \to \infty} (x + 1)/(x^2 + 1)
  10. Ex. 42.10Application
    limx(x3+1)/(x21)\lim_{x \to -\infty} (x^3 + 1)/(x^2 - 1)。(答:-\infty。)
  11. Ex. 42.11Application
    limx0x2sin(1/x)\lim_{x \to 0} x^2 \sin(1/x),用夹逼。
  12. Ex. 42.12ApplicationAnswer key
    limx0(ex1)/x\lim_{x \to 0} (e^x - 1)/x
  13. Ex. 42.13Application
    limx0ln(1+x)/x\lim_{x \to 0} \ln(1+x)/x
  14. Ex. 42.14ApplicationAnswer key
    limx1lnx/(x1)\lim_{x \to 1} \ln x / (x - 1)
  15. Ex. 42.15Application
    limx0(1+2x)1/x\lim_{x \to 0} (1 + 2x)^{1/x}。(答:e2e^2。)
  16. Ex. 42.16Application
    limx4(x4)/(x2)\lim_{x \to 4} (x - 4)/(\sqrt x - 2)
  17. Ex. 42.17Application
    limx(sinx)/x\lim_{x \to \infty} (\sin x)/x,用夹逼。
  18. Ex. 42.18ApplicationAnswer key
    limn(2n2+5n)/(n2+1)\lim_{n \to \infty} (2n^2 + 5n)/(n^2 + 1)
  19. Ex. 42.19Application
    limx0(cosx)1/x2\lim_{x \to 0} (\cos x)^{1/x^2}。(答:e1/2e^{-1/2}。)
  20. Ex. 42.20ApplicationAnswer key
    limx0(1+x1x)/x\lim_{x \to 0} (\sqrt{1+x} - \sqrt{1-x})/x。(答:1。)
  21. Ex. 42.21Understanding
    用夹逼定理 + 单位圆几何证 limx0sinx/x=1\lim_{x \to 0} \sin x / x = 1
  22. Ex. 42.22Understanding
    用夹逼证 limx0xcos(1/x)=0\lim_{x \to 0} x \cos(1/x) = 0
  23. Ex. 42.23Understanding
    limx0+xx=1\lim_{x \to 0^+} x^x = 1。(用 xx=exlnxx^x = e^{x \ln x}limxlnx=0\lim x \ln x = 0。)
  24. Ex. 42.24Understanding
    lim(f+g)\lim (f + g) 存在但 limf\lim flimg\lim g 不存在的例。
  25. Ex. 42.25Understanding
    lim(fg)\lim (f \cdot g) 存在但 limf,limg\lim f, \lim g 只一个存在的例。
  26. Ex. 42.26Understanding
    证若 limf(x)=0\lim |f(x)| = 0,则 limf(x)=0\lim f(x) = 0
  27. Ex. 42.27Proof
    用 ε-δ 证 lim(f+g)=limf+limg\lim (f + g) = \lim f + \lim g
  28. Ex. 42.28ProofAnswer key
    用 ε-δ 证 lim(cf)=climf\lim (cf) = c \lim f
  29. Ex. 42.29Proof
    用 ε-δ 证夹逼定理。
  30. Ex. 42.30Proof
    证若 limf=L\lim f = LL>0L > 0,存在 f>L/2f > L/2 的邻域。
  31. Ex. 42.31Proof
    证极限(存在时)唯一。
  32. Ex. 42.32ProofAnswer key
    证复合:limxag=b\lim_{x \to a} g = bffbb 连续 ⇒ limf(g(x))=f(b)\lim f(g(x)) = f(b)
  33. Ex. 42.33Modeling
    在控制中,传递函数 H(s)=(s+2)/(s2+3s+2)H(s) = (s+2)/(s^2 + 3s + 2)。用性质算 H(0)H(0)
  34. Ex. 42.34ModelingAnswer key
    在药动学中,C(t)=(D/V)(ekateket)/(keka)C(t) = (D/V)(e^{-k_a t} - e^{-k_e t})/(k_e - k_a)。计算 limtC(t)\lim_{t \to \infty} C(t)
  35. Ex. 42.35Modeling
    2\sqrt 2 的牛顿迭代序列:an+1=(an+2/an)/2a_{n+1} = (a_n + 2/a_n)/2。用性质找极限。
  36. Ex. 42.36Modeling
    信号 V(t)=Asin(ωt)/tV(t) = A \sin(\omega t)/ttt \to \infty 极限?t0t \to 0
  37. Ex. 42.37Modeling
    在概率中,XnXX_n \to X 分布收敛。用 cos\cos 连续性证 E[cosXn]E[cosX]E[\cos X_n] \to E[\cos X]
  38. Ex. 42.38Modeling
    欧拉法误差:y(xn)ynCh|y(x_n) - y_n| \leq C hh0h \to 0。建模为极限。
  39. Ex. 42.39Challenge
    limx0(sinxx)/x3\lim_{x \to 0} (\sin x - x)/x^3。(答:1/6-1/6。)
  40. Ex. 42.40Challenge
    limx0((1+x)1/xe)/x\lim_{x \to 0} ((1+x)^{1/x} - e)/x
  41. Ex. 42.41Challenge
    limxxsin(π/x)\lim_{x \to \infty} x \sin(\pi/x)。(答:π\pi。)
  42. Ex. 42.42Challenge
    limx0(cosxcos(2x))/x2\lim_{x \to 0} (\cos x - \cos(2x))/x^2。(答:3/23/2。)

参考来源

Updated on 2026-04-30 · Author(s): Clube da Matemática

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