Limits of Rational Functions: Public Worked Question

Question

Evaluate $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6}$ .

C. $6$

1Direct substitution gives $\dfrac{9 - 9}{9 - 15 + 6} = \dfrac{0}{0}$ , an indeterminate form, so the limit laws cannot be applied directly.
2Factor both polynomials: $x^2 - 9 = (x-3)(x+3)$ and $x^2 - 5x + 6 = (x-3)(x-2)$ .
3Cancel the common factor $(x-3)$ , valid for $x \neq 3$ , leaving $\dfrac{x+3}{x-2}$ .
4Substitute $x = 3$ into the simplified expression: $\dfrac{3+3}{3-2} = 6$ .

This substitutes $x = 3$ into the numerator alone and concludes the limit is zero, ignoring that the denominator also vanishes there.

This takes the ratio of the leading coefficients, which is the behavior as $x \to \infty$ , not the limit at the finite point $x = 3$ .

Correct. Factoring gives $\dfrac{(x-3)(x+3)}{(x-3)(x-2)} = \dfrac{x+3}{x-2}$ for $x \neq 3$ , and substituting yields $\dfrac{6}{1} = 6$ .

This stops at the indeterminate form $0/0$ , which signals more work is needed rather than proving the limit fails to exist.

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{F(x)}{G(x)}, \quad f = (x-a)F, \; g = (x-a)G

Assumptions: The simplified denominator G is nonzero at a, so the reduced expression can be evaluated by substitution.

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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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