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第41课 — 正式极限:ε-δ 定义

A definição ε-δ de limite. Cauchy 1821, Weierstrass 1872. O ponto onde o cálculo se torna rigoroso.

Used in: 2.º ano EM (16-17 anos) · Equiv. Math II japonês · Equiv. Klasse 11 alemã (Analysis) · A-Level Further Maths — Limits

limxaf(x)=L    ε>0,δ>0:0<xa<δf(x)L<ε\lim_{x \to a} f(x) = L \iff \forall \varepsilon > 0, \exists \delta > 0 : 0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon
Choose your door

Rigorous notation, full derivation, hypotheses

严格定义

通过 ε-δ 证明极限

  1. xa|x - a| 表示 f(x)L|f(x) - L|
  2. 任意 \eps>0\eps > 0,找 δ=δ(\eps)\delta = \delta(\eps) 使 0<xa<δ0 < |x - a| < \deltaf(x)L<\eps|f(x) - L| < \eps

模板例:limx2(3x+1)=7\lim_{x \to 2} (3x + 1) = 7

3x+17=3x6=3x2|3x + 1 - 7| = |3x - 6| = 3|x - 2|。给定 \eps>0\eps > 0,取 δ=\eps/3\delta = \eps/3。则 x2<δ3x2<\eps|x - 2| < \delta \Rightarrow 3|x - 2| < \eps。∎

单侧极限

  • limxa+f(x)=L\lim_{x \to a^+} f(x) = L0<xa<δ0 < x - a < \delta(仅右侧)。
  • limxaf(x)=L\lim_{x \to a^-} f(x) = L0<ax<δ0 < a - x < \delta(仅左侧)。
  • limxaf(x)=L    limxa+=limxa=L\lim_{x \to a} f(x) = L \iff \lim_{x \to a^+} = \lim_{x \to a^-} = L

无穷极限

limxaf(x)=+\lim_{x \to a} f(x) = +\inftyM>0,δ>0:0<xa<δf(x)>M\forall M > 0, \exists \delta > 0 : 0 < |x - a| < \delta \Rightarrow f(x) > M

limxf(x)=L\lim_{x \to \infty} f(x) = L\eps>0,N:x>Nf(x)L<\eps\forall \eps > 0, \exists N : x > N \Rightarrow |f(x) - L| < \eps

4 种量化表

类型量化
limxa=L\lim_{x \to a} = L\eps,δ\forall \eps, \exists \delta
limxa=\lim_{x \to a} = \inftyM,δ\forall M, \exists \delta
limx=L\lim_{x \to \infty} = L\eps,N\forall \eps, \exists N
limx=\lim_{x \to \infty} = \inftyM,N\forall M, \exists N

基本性质

  • 和的极限 = 极限的和。
  • 积的极限 = 极限的积。
  • 商的极限 = 商,分母 0\neq 0

Exercise list

40 exercises · 10 with worked solution (25%)

Application 20Understanding 3Modeling 5Challenge 5Proof 7
  1. Ex. 41.1Application
    limx3(2x+1)\lim_{x \to 3} (2x + 1)。(答:7。)
  2. Ex. 41.2Application
    limx0(x2+4x+7)\lim_{x \to 0} (x^2 + 4x + 7)
  3. Ex. 41.3ApplicationAnswer key
    limx2(x24)/(x2)\lim_{x \to 2} (x^2 - 4)/(x - 2)。(答:4。)
  4. Ex. 41.4Application
    limx1(x21)/(x1)\lim_{x \to 1} (x^2 - 1)/(x - 1)
  5. Ex. 41.5ApplicationAnswer key
    limx0(x+11)/x\lim_{x \to 0} (\sqrt{x+1} - 1)/x。(乘共轭。)
  6. Ex. 41.6ApplicationAnswer key
    limx(3x+1)/(x+5)\lim_{x \to \infty} (3x + 1)/(x + 5)。(答:3。)
  7. Ex. 41.7Application
    limx(2x2+3)/(x21)\lim_{x \to \infty} (2x^2 + 3)/(x^2 - 1)
  8. Ex. 41.8Application
    limx0sinx/x\lim_{x \to 0} \sin x / x
  9. Ex. 41.9Application
    limx0sin(2x)/x\lim_{x \to 0} \sin(2x)/x。(答:2。)
  10. Ex. 41.10Application
    limx0(1cosx)/x2\lim_{x \to 0} (1 - \cos x)/x^2。(答:1/21/2。)
  11. Ex. 41.11Application
    limx(1+1/x)x\lim_{x \to \infty} (1 + 1/x)^x
  12. Ex. 41.12ApplicationAnswer key
    limx0(ex1)/x\lim_{x \to 0} (e^x - 1)/x
  13. Ex. 41.13Application
    limx0+1/x\lim_{x \to 0^+} 1/x
  14. Ex. 41.14ApplicationAnswer key
    limx01/x\lim_{x \to 0^-} 1/x
  15. Ex. 41.15ApplicationAnswer key
    limx2(x24)/(x25x+6)\lim_{x \to 2} (x^2 - 4)/(x^2 - 5x + 6)。(答:4-4。)
  16. Ex. 41.16Application
    limx0sin(5x)/sin(3x)\lim_{x \to 0} \sin(5x)/\sin(3x)
  17. Ex. 41.17Application
    limxx2+1x\lim_{x \to \infty} \sqrt{x^2 + 1} - x。(答:0。)
  18. Ex. 41.18Application
    limx4(x2)/(x4)\lim_{x \to 4} (\sqrt x - 2)/(x - 4)
  19. Ex. 41.19Application
    limx0sin(x2)/x\lim_{x \to 0} \sin(x^2)/x
  20. Ex. 41.20Application
    limx(lnx)/x\lim_{x \to \infty} (\ln x)/x。(答:0。)
  21. Ex. 41.21ProofAnswer key
    用 ε-δ 证 limx5(2x+3)=13\lim_{x \to 5} (2x + 3) = 13
  22. Ex. 41.22Proof
    用 ε-δ 证 limx3x2=9\lim_{x \to 3} x^2 = 9
  23. Ex. 41.23Proof
    limx01/x\lim_{x \to 0} 1/x 不存在。
  24. Ex. 41.24Proof
    limx0sin(1/x)\lim_{x \to 0} \sin(1/x) 不存在。
  25. Ex. 41.25ProofAnswer key
    用 ε-δ 证 limx1(x3)=1\lim_{x \to 1} (x^3) = 1
  26. Ex. 41.26ProofAnswer key
    证若极限存在,唯一。
  27. Ex. 41.27Proof
    证夹逼定理(三明治)。
  28. Ex. 41.28Understanding
    证若 limxaf(x)=L\lim_{x \to a} f(x) = Llimxag(x)=M\lim_{x \to a} g(x) = M,则 lim(f+g)=L+M\lim (f + g) = L + M
  29. Ex. 41.29Understanding
    说明为何 f(a)f(a) 不必定义而 limxaf(x)\lim_{x \to a} f(x) 存在。给例。
  30. Ex. 41.30Understanding
    构造 ff 使 limx0+f=1\lim_{x \to 0^+} f = 1limx0f=1\lim_{x \to 0^-} f = -1limx0f\lim_{x \to 0} f 存在?
  31. Ex. 41.31Modeling
    RC 电路中,V(t)=V(1et/τ)V(t) = V_\infty (1 - e^{-t/\tau})。计算 limtV(t)\lim_{t \to \infty} V(t) 并解释。
  32. Ex. 41.32Modeling
    s(t)=t2s(t) = t^2 的瞬时速度 v(t)=limΔt0(s(t+Δt)s(t))/Δtv(t) = \lim_{\Delta t \to 0} (s(t + \Delta t) - s(t))/\Delta t
  33. Ex. 41.33Modeling
    在药动学中,C(t)=C0ektC(t) = C_0 e^{-kt}。计算 limtC(t)\lim_{t \to \infty} C(t)
  34. Ex. 41.34Modeling
    在控制中,传递函数 H(s)=K/(s+1)H(s) = K/(s+1)。计算 lims0H(s)\lim_{s \to 0} H(s)(DC 增益)。
  35. Ex. 41.35Modeling
    泰勒截断误差:limh0(f(x+h)f(x)hf(x))/h2=f(x)/2\lim_{h \to 0} (f(x+h) - f(x) - hf'(x))/h^2 = f''(x)/2。对 f(x)=exf(x) = e^xx=0x = 0 验证。
  36. Ex. 41.36Challenge
    limx0(tanxx)/x3\lim_{x \to 0} (\tan x - x)/x^3。(答:1/31/3。)
  37. Ex. 41.37Challenge
    limx(x2+xx)\lim_{x \to \infty} (\sqrt{x^2 + x} - x)。(答:1/21/2。)
  38. Ex. 41.38Challenge
    limx0ex1xx2\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}
  39. Ex. 41.39Challenge
    limxπ/2(1sinx)secx\lim_{x \to \pi/2} (1 - \sin x)^{\sec x}
  40. Ex. 41.40ChallengeAnswer key
    用 ε-δ 证 limx21/x=1/2\lim_{x \to 2} 1/x = 1/2

参考来源

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

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