Limits by Factoring and Cancelling - Calculus Step by Step

Evaluate a 0/0 indeterminate form by factoring the numerator and cancelling the common factor.

Example

Resolve a 0 over 0 limit by factoring and canceling a common factor. The limit exists even though the function is undefined at that point; the theorem that makes this precise is the epsilon-delta definition of a limit, which this book uses but does not prove.

highlighted = computed this step

Step 1 — Set up

Set up the factoring limit at x = 3.

\lim_{x\to 3 } \frac{x^{2}-9}{x-3}

Step 2 — Check the form

Direct substitution gives 0 over 0.

\hlmath{\frac{0}{0}}

Step 3 — Factor numerator

Factor the numerator as x plus 3 times x minus 3.

\frac{ \hlmath{(x+3)(x-3)} }{x- 3 }

Step 4 — Cancel common factor

Cancel the common x minus 3 factor.

\hlmath{x+3}

Step 5 — Substitute

Substitute 3 into x plus 3 to get 6.

3 + 3 = \hl{6}

limit-factor-cancel When direct substitution gives 0/0, factor the numerator to reveal a common factor with the denominator. Cancel the common factor, then substitute.