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第48课 — 三角函数极限

Manipulação de limites trig. Identidades aplicadas a sin x/x, 1−cos x, tan x e variantes.

Used in: 2.º ano do EM (16 anos) · Equiv. Math II japonês cap. 4 · Equiv. Klasse 11 alemã (Grenzwerte trigonometrischer Funktionen)

limx0sin(kx)x=k,limx01cosxx2=12\lim_{x \to 0} \frac{\sin(kx)}{x} = k, \quad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}
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Rigorous notation, full derivation, hypotheses

三角操作

基本三角极限

极限
limx0sinx/x\lim_{x \to 0} \sin x / x11
limx0sin(kx)/x\lim_{x \to 0} \sin(kx)/xkk
limx0tanx/x\lim_{x \to 0} \tan x / x11
limx0(1cosx)/x\lim_{x \to 0} (1 - \cos x)/x00
limx0(1cosx)/x2\lim_{x \to 0} (1 - \cos x)/x^21/21/2
limx0sin2x/x2\lim_{x \to 0} \sin^2 x / x^211
limx0arcsinx/x\lim_{x \to 0} \arcsin x / x11
limx0arctanx/x\lim_{x \to 0} \arctan x / x11

有用恒等式

  • 1cosx=2sin2(x/2)1 - \cos x = 2\sin^2(x/2)
  • sin(a+b)=sinacosb+cosasinb\sin(a + b) = \sin a \cos b + \cos a \sin b
  • cos(a+b)=cosacosbsinasinb\cos(a + b) = \cos a \cos b - \sin a \sin b
  • 1cosx=(1cos2x)/(1+cosx)=sin2x/(1+cosx)1 - \cos x = (1 - \cos^2 x)/(1 + \cos x) = \sin^2 x / (1 + \cos x)

技巧

  1. 替换 uusin(2x)/x=2sin(2x)/(2x)2\sin(2x)/x = 2 \cdot \sin(2x)/(2x) \to 2
  2. 恒等式重写为基本形式。
  3. 共轭1cosx=sin2x/(1+cosx)1 - \cos x = \sin^2 x / (1 + \cos x)
  4. 夹逼1sinx1-1 \leq \sin x \leq 1,乘以 0\to 0 的东西。

lim(1cosx)/x2=1/2\lim (1 - \cos x)/x^2 = 1/2 证明

1cosxx2=2sin2(x/2)x2=12sin2(x/2)(x/2)2121=12\frac{1 - \cos x}{x^2} = \frac{2 \sin^2(x/2)}{x^2} = \frac{1}{2} \cdot \frac{\sin^2(x/2)}{(x/2)^2} \to \frac{1}{2} \cdot 1 = \frac{1}{2}

limtanx/x=1\lim \tan x / x = 1 证明

tanx/x=(sinx/x)(1/cosx)11=1\tan x / x = (\sin x / x) \cdot (1 / \cos x) \to 1 \cdot 1 = 1

Exercise list

40 exercises · 10 with worked solution (25%)

Application 30Understanding 2Modeling 6Challenge 2
  1. Ex. 48.1Application
    limx0sin(7x)/x\lim_{x \to 0} \sin(7x)/x。(答:7。)
  2. Ex. 48.2Application
    limx0sin(3x)/sin(5x)\lim_{x \to 0} \sin(3x)/\sin(5x)。(答:3/53/5。)
  3. Ex. 48.3Application
    limx0tan(2x)/sin(3x)\lim_{x \to 0} \tan(2x)/\sin(3x)。(答:2/32/3。)
  4. Ex. 48.4Application
    limx0sin2x/x\lim_{x \to 0} \sin^2 x / x。(答:0。)
  5. Ex. 48.5ApplicationAnswer key
    limx0(1cos(2x))/x2\lim_{x \to 0} (1 - \cos(2x))/x^2。(答:2。)
  6. Ex. 48.6Application
    limxπ/2(1sinx)/(π/2x)2\lim_{x \to \pi/2} (1 - \sin x)/(\pi/2 - x)^2。(答:1/21/2。)
  7. Ex. 48.7Application
    limx0(tanxsinx)/x3\lim_{x \to 0} (\tan x - \sin x)/x^3。(答:1/21/2。)
  8. Ex. 48.8ApplicationAnswer key
    limx0arcsin(2x)/x\lim_{x \to 0} \arcsin(2x)/x。(答:2。)
  9. Ex. 48.9Application
    limx0arctan(3x)/sin(2x)\lim_{x \to 0} \arctan(3x)/\sin(2x)。(答:3/23/2。)
  10. Ex. 48.10Application
    limxπsinx/(xπ)\lim_{x \to \pi} \sin x/(x - \pi)。(答:1-1。)
  11. Ex. 48.11Application
    limx0(cosx1)/(sinx)\lim_{x \to 0} (\cos x - 1)/(\sin x)。(答:0。)
  12. Ex. 48.12Application
    limx0xcotx\lim_{x \to 0} x \cdot \cot x。(答:1。)
  13. Ex. 48.13Application
    limx0sin(x3)/x\lim_{x \to 0} \sin(x^3)/x。(答:0。)
  14. Ex. 48.14ApplicationAnswer key
    limx0sin(x)sin(2x)/x2\lim_{x \to 0} \sin(x)\sin(2x)/x^2。(答:2。)
  15. Ex. 48.15ApplicationAnswer key
    limx0(1cos(3x))/(1cos(2x))\lim_{x \to 0} (1 - \cos(3x))/(1 - \cos(2x))。(答:9/49/4。)
  16. Ex. 48.16Application
    limx0(1cosx)/x\lim_{x \to 0} (1 - \cos x)/x。(答:0。)
  17. Ex. 48.17ApplicationAnswer key
    limx0sin(5x)cot(3x)\lim_{x \to 0} \sin(5x) \cdot \cot(3x)。(答:5/35/3。)
  18. Ex. 48.18ApplicationAnswer key
    limx0(cosxcos(3x))/x2\lim_{x \to 0} (\cos x - \cos(3x))/x^2。(答:4。)
  19. Ex. 48.19Application
    limx0sin(πx)/sin(π(1x))\lim_{x \to 0} \sin(\pi x)/\sin(\pi(1-x))
  20. Ex. 48.20Application
    limx0sin(sinx)/x\lim_{x \to 0} \sin(\sin x)/x。(答:1。)
  21. Ex. 48.21Application
    limxπ/4(tanx1)/(xπ/4)\lim_{x \to \pi/4} (\tan x - 1)/(x - \pi/4)。(答:2。)
  22. Ex. 48.22ApplicationAnswer key
    limxπ/2cosx/(xπ/2)\lim_{x \to \pi/2} \cos x / (x - \pi/2)。(答:1-1。)
  23. Ex. 48.23Application
    limx0(secx1)/x2\lim_{x \to 0} (\sec x - 1)/x^2。(答:1/21/2。)
  24. Ex. 48.24Application
    limx0(tan2x)/x2\lim_{x \to 0} (\tan^2 x)/x^2。(答:1。)
  25. Ex. 48.25Application
    limx0(1cosxcos(2x))/x2\lim_{x \to 0} (1 - \cos x \cdot \cos(2x))/x^2。(答:5/25/2。)
  26. Ex. 48.26ApplicationAnswer key
    limx0sin(πx)/x\lim_{x \to 0} \sin(\pi - x)/x
  27. Ex. 48.27Application
    limx0(sin(a+x)sina)/x\lim_{x \to 0} (\sin(a+x) - \sin a)/x。(答:cosa\cos a。)
  28. Ex. 48.28Application
    limx0(cos(a+x)cosa)/x\lim_{x \to 0} (\cos(a+x) - \cos a)/x。(答:sina-\sin a。)
  29. Ex. 48.29Application
    limx0sinxln(1+x)/x2\lim_{x \to 0} \sin x \cdot \ln(1+x)/x^2。(答:1。)
  30. Ex. 48.30Application
    limx0(excosx)/x\lim_{x \to 0} (e^x - \cos x)/x。(答:1。)
  31. Ex. 48.31ModelingAnswer key
    控制中,H(s)H(s)jωj\omega 极点——共振时 \to \infty 极限。
  32. Ex. 48.32Modeling
    小角度 sinxxx3/6\sin x \approx x - x^3/6 近似。x=0,1x = 0{,}1 相对误差?
  33. Ex. 48.33Modeling
    力学中钟摆:θ¨=(g/L)sinθ(g/L)θ\ddot \theta = -(g/L) \sin \theta \approx -(g/L)\theta 小角度。通过极限证。
  34. Ex. 48.34Modeling
    折射:n1sinθ1=n2sinθ2n_1 \sin \theta_1 = n_2 \sin \theta_2。傍轴近似 n1θ1n2θ2n_1 \theta_1 \approx n_2 \theta_2
  35. Ex. 48.35Modeling
    AM 信号 V(t)=(1+mcos(ωmt))cos(ωct)V(t) = (1 + m \cos(\omega_m t)) \cos(\omega_c t)。小 mm 通过极限展开。
  36. Ex. 48.36Modeling
    单缝衍射:(sinu/u)2(\sin u/u)^2 模式,u=πasinθ/λu = \pi a \sin\theta/\lambdaθ=0\theta = 0 极限。
  37. Ex. 48.37Understanding
    通过变量替换证 limx0sin(kx)/x=k\lim_{x \to 0} \sin(kx)/x = k
  38. Ex. 48.38Understanding
    用恒等式 1cosx=2sin2(x/2)1 - \cos x = 2\sin^2(x/2)(1cosx)/x21/2(1 - \cos x)/x^2 \to 1/2
  39. Ex. 48.39Challenge
    limx0(sin(sinx)x)/x3\lim_{x \to 0} (\sin(\sin x) - x)/x^3
  40. Ex. 48.40ChallengeAnswer key
    limx0(cos(sinx)cosx)/x4\lim_{x \to 0} (\cos(\sin x) - \cos x)/x^4

参考来源

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

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