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EasyLimits and ContinuityCalculus0:53AP Calculus AB, unit 1

Evaluating a Limit by Factoring

Direct substitution gives 0/0 for the limit of (x² − 9)/(x − 3) as x → 3. Factor, cancel and substitute to get 6, and see the hole on the graph.

Narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)). Download MP4 (0.6 MB) · Captions (WebVTT)

On-screen maths

Every board from the lesson, so you can read it at your own pace or copy it into your notes.

  1. 1. Try substitution0:00
    1. lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}
    2. 32−93−3=00\frac{3^2 - 9}{3 - 3} = \frac{0}{0}
      indeterminate: more work needed
  2. 2. Factor and cancel0:14
    1. x2−9x−3=(x−3)(x+3)x−3\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3}
    2. =x+3(x≠3)= x + 3 \quad (x \ne 3)
    3. lim⁡x→3(x+3)=6\lim_{x \to 3} (x + 3) = 6
  3. 3. The hole in the graph0:33
    -1123456-112345678910hole at (3, 6)
    y=x2−9x−3y = \frac{x^2 - 9}{x - 3}
  4. 4. Recap0:45
    • 00\frac{0}{0} means: simplify first.
    • Factor, cancel the common factor, substitute.
    • The limit can exist where the function does not.

About limits and continuity

Evaluate limits algebraically and decide continuity.

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