Hole or Vertical Asymptote? Factor a Rational Function Before Deciding

A cancelled denominator factor can create a hole if the remaining expression has a finite limit there. An uncancelled denominator zero may create a vertical asymptote; check the nearby behaviour.

Both begin with an excluded input

A rational expression is undefined where its original denominator is zero. That restriction survives algebraic cancellation. However, different denominator zeros can produce different graph behaviour. At a removable hole, nearby values approach a finite height. At a vertical asymptote, at least one one-sided limit grows without bound.

Factoring is the starting point, not the final verdict. Simplify only on the original domain, then inspect each excluded value in the remaining expression. Record the hole’s coordinate as well as its x-value when the problem requests a graph feature.

Worked example with both features

Let f(x) = (x² − 9)/[(x − 3)(x − 1)]. Its original domain excludes x = 3 and x = 1. Factor the numerator as (x − 3)(x + 3). For permitted inputs, the function equals (x + 3)/(x − 1).

At x = 3, this remaining expression has finite value 6/2 = 3. Thus the original graph has a hole at (3, 3), rather than a point actually included in the graph. At x = 1, the remaining numerator approaches 4 while its denominator approaches zero. From the right the quotient tends to positive infinity; from the left it tends to negative infinity. The line x = 1 is a vertical asymptote.

Multiplicity matters

If a numerator has fewer copies of a factor than the denominator, cancellation can leave a denominator copy behind. For example, (x − 2)/(x − 2)² simplifies to 1/(x − 2), still with a vertical asymptote at 2. Seeing a shared factor does not automatically mean the zero is removable.

Conversely, (x − 2)²/(x − 2) simplifies to x − 2 for x ≠ 2. The remaining limit is zero, so the original graph has a hole at (2, 0). Counting factor multiplicities prevents these two cases from being confused.

Common mistake

Do not call every original denominator zero an asymptote. Also do not fill a hole by silently replacing the original function with its simplified expression at an excluded input. A new definition could fill that hole, but it would be a modified function.

Use three separate lines in your answer: original domain, simplified rule on that domain, and behaviour at each excluded input. This makes it clear which conclusions came from algebra and which came from a limit.

Cancel hua factor aur bacha denominator alag dekho—dono ka graph alag ho sakta hai.

Three practice questions with worked answers

1. Classify x = 2 for (x² − 4)/(x − 2).

   For x ≠ 2, the expression equals x + 2. Its limit at 2 is 4, so there is a hole at (2, 4), not a vertical asymptote.

2. Classify x = 4 for 1/(x − 4).

   There is no cancelling numerator factor. The quotient grows without bound near 4, so x = 4 is a vertical asymptote.

3. Classify x = 1 for (x − 1)²/(x − 1).

   The original excludes 1, but simplifies to x − 1 elsewhere. The finite limit is zero, giving a hole at (1, 0).

Related study guides

• Domain and Range from a Function Rule
• Limits and Derivatives Class 11: Standard Limits and Solved Examples

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Limits at Infinity and Asymptotes. No textbook or exam-board affiliation is claimed.