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A limit at a missing point

Factor before cancelling, retain the domain restriction, and distinguish a limit from a function value.

Question

For x≠3x\ne3, let f(x)=(x2−9)/(x−3)f(x)=(x^2-9)/(x-3). What is lim⁡x→3f(x)\lim_{x\to3} f(x)?

  1. a.0
  2. b.3
  3. c.6
  4. d.The limit does not exist.

Correct answer

C. 6

Full reasoning

  1. 1Factor x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3).
  2. 2For x≠3x\ne3, cancelling the nonzero factor gives f(x)=x+3f(x)=x+3.
  3. 3The limit depends on nearby inputs, so take the limit of x + 3 to obtain 6.
  4. 4Do not claim the original formula is defined at 3. Defining f(3) = 6 would create a continuous extension; defining a different value would leave the same limit but not continuity.

Why each choice is right or wrong

Choice A

A zero numerator by itself does not determine a limit when the denominator also tends to zero.

Choice B

After cancelling the common factor, the remaining expression is x + 3, whose nearby values approach 6 rather than 3.

Choice C

Correct. Away from x = 3, f(x) equals x + 3, whose limit is 6.

Choice D

The expression being undefined at the point does not stop nearby values from approaching a limit.

Related formula

Difference of squares

x² − a² = (x − a)(x + a)

Assumptions: Cancellation by x − a additionally requires x ≠ a.

Related topic and practice

Calculus Domain GuideSeries Convergence TestsGeometric Series ConvergenceLimits and continuityCreate a free account to practice

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Sources

Original authored exercise; no database question ID and no protected or official exam-bank content. Specialist review pending.

  1. Calculus Volume 1, Section 2.3: The Limit Laws — OpenStax. Accessed 2026-08-03. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
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Sources last checked 2026-09-13

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