A 0/0 Limit by Factoring: Cancel for Nearby Inputs, Then Take the Limit

If direct substitution produces 0/0, it has not evaluated the limit. Factor, cancel a common factor for nearby permitted inputs, and take the limit of the simplified expression.

Why 0/0 is not an answer

The form 0/0 is indeterminate: different functions giving that form can have different limits. For example, x/x approaches 1 near zero, while 2x/x approaches 2. The substitution pattern alone cannot decide which answer is correct.

A limit concerns values arbitrarily near the target, not necessarily the value at the target. If two expressions agree throughout a punctured neighbourhood, they have the same limit there whenever that limit exists. This is what allows cancellation in a limit without claiming division by zero is valid.

Worked example

Find lim as x → 5 of (x² − 25)/(x − 5). Direct substitution gives 0/0. Factor x² − 25 = (x − 5)(x + 5). For x ≠ 5, the fraction equals x + 5. The limit is therefore 5 + 5 = 10.

The original fraction remains undefined at x = 5. We have found its nearby limiting value, not assigned it a value there. A table illustrates the approach: x = 4.9 gives 9.9, x = 4.99 gives 9.99, x = 5.01 gives 10.01, and x = 5.1 gives 10.1.

A cube factor example

For lim as x → 2 of (x³ − 8)/(x − 2), factor the numerator into (x − 2)(x² + 2x + 4). Cancelling for x ≠ 2 leaves a polynomial whose limit is 4 + 4 + 4 = 12. Recognising a difference of cubes keeps the calculation short.

The same idea does not justify cancelling separate terms across addition. In (x² + 1)/(x + 1), the x terms do not form a common multiplicative factor. Factor the whole numerator and denominator before cancelling.

When factoring does not settle the limit

A nonzero numerator over a denominator approaching zero usually requires a one-sided sign analysis rather than this cancellation method. Other 0/0 forms may need rationalisation, a trigonometric identity or another justified tool. Identify the actual structure instead of forcing every limit into one recipe.

Common mistake

Do not write “cancel x − 5 at x = 5.” Cancellation is valid for x ≠ 5, precisely where the limit investigates nearby values. Keep that qualification in the working, and only substitute the target after the simplified expression is continuous there.

0/0 answer nahi; pehle expression ko simplify karne ka signal hai.

Three practice questions with worked answers

1. Find lim as x → 3 of (x² − 9)/(x − 3).

   For x ≠ 3, the quotient is x + 3. Its limit is 3 + 3 = 6; the original quotient is still undefined at 3.

2. Find lim as x → 4 of (x² − 16)/(x − 4).

   Factor the numerator as (x − 4)(x + 4). The nearby expression is x + 4, so the limit is 8.

3. Find lim as x → 1 of (x³ − 1)/(x − 1).

   The quotient equals x² + x + 1 for x ≠ 1. Substitution into that polynomial gives 1 + 1 + 1 = 3.

Related study guides

• Limits and Derivatives Class 11: Standard Limits and Solved Examples
• Factor Theorem and Zero Check

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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: The Limit Laws. No textbook or exam-board affiliation is claimed.