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LimitsCalculusCollegeabout 2:18

Find lim x->0 (1-cos(3x))/x^2: animated lesson

A narrated, step-by-step limits video lesson with captions. Final answer: Limit = 9/2 ≈ 4.5.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.9 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    lim⁡x→01−cos⁡(3x)x2\lim_{x \to 0} \frac{1-\cos\left(3x\right)}{{x}^{2}}
  2. 2. Step by step
    1. lim⁡x→01−cos⁡(3x)x2=lim⁡x→03sin⁡(3x)2x\lim_{x \to 0} \frac{1-\cos\left(3x\right)}{{x}^{2}} = \lim_{x \to 0} \frac{3\sin\left(3x\right)}{2x}
      Indeterminate form 0/0 — apply L'Hôpital's rule
    2. lim⁡x→03sin⁡(3x)2x=lim⁡x→09cos⁡(3x)2\lim_{x \to 0} \frac{3\sin\left(3x\right)}{2x} = \lim_{x \to 0} \frac{9\cos\left(3x\right)}{2}
      Indeterminate form 0/0 — apply L'Hôpital's rule
    3. lim⁡x→09cos⁡(3x)2=9cos⁡(3(0))2=92≈4.5\lim_{x \to 0} \frac{9\cos\left(3x\right)}{2} = \frac{9\cos\left(3\left(0\right)\right)}{2} = \frac{9}{2} \approx 4.5
      Direct substitution
  3. 3. The answer
    lim⁡x→01−cos⁡(3x)x2=92≈4.5\lim_{x \to 0} \frac{1-\cos\left(3x\right)}{{x}^{2}} = \frac{9}{2} \approx 4.5

    lim x→0 (1 - cos(3x)) / x² = 9/2 ≈ 4.5

  4. 4. Check
    • Numerical values from the right agree: f(0.01) = 4.4996625, f(0.001) = 4.4999966, f(0.0001) = 4.5, f(0.00001) = 4.5000004 ✓
    • Numerical values from the left agree: f(-0.01) = 4.4996625, f(-0.001) = 4.4999966, f(-0.0001) = 4.5, f(-0.00001) = 4.5000004 ✓

Transcript

The problem. Here is the problem: (1-cos(3x))/x squared. Find the limit of (1 minus cos(3x)) / x squared as x → 0.

Step by step. Indeterminate form 0 over 0 — apply L prime Hôpital prime s rule. Differentiate the numerator and the denominator separately (this is not the quotient rule). Indeterminate form 0 over 0 — apply L prime Hôpital prime s rule. Differentiate the numerator and the denominator separately (this is not the quotient rule). Direct substitution. The expression is continuous at x equals 0, so the limit equals its value there.

The answer. So the answer is Limit equals 9 over 2 approximately 4.5. A limit describes the value a function approaches, not necessarily the value it takes. When substitution gives an indeterminate form, the expression is rewritten (L prime Hôpital prime s rule, a conjugate, or logarithms) until substitution works.

Check. Numerical values from the right agree: f(0.01) equals 4.4996625, f(0.001) equals 4.4999966, f(0.0001) equals 4.5, f(0.00001) equals 4.5000004, which checks out. Numerical values from the left agree: f(negative 0.01) equals 4.4996625, f(negative 0.001) equals 4.4999966, f(negative 0.0001) equals 4.5, f(negative 0.00001) equals 4.5000004, which checks out.

About limits

Evaluate limits by substitution, factoring, standard limits and L'Hôpital's rule.

Find lim x->0 (1-cos(3x))/x^2: Limits Step-by-Step Lesson