Resolving 0/0 Limits Algebraically

When direct substitution hands you the indeterminate form \(0/0\), the limit isn't dead — it's hiding. Rewrite the expression into an equivalent one that IS defined at the target point: factor and cancel, rationalize with a conjugate, or simplify a complex fraction.

By the end you'll be able to recognize a \(0/0\) form, pick the right algebraic move for the expression in front of you — factoring, conjugate, or fraction cleanup — and evaluate the limit the quotient law couldn't touch.

Predict: \(h(x) = \dfrac{x^2-9}{x-3}\) is undefined at \(x = 3\) — substitution gives \(0/0\). What value does the curve approach at the hole? Drag x toward 3 to test your guess, then hit Factor & cancel to check it against the simplified line \(y = x + 3\).

h(x) = (x² − 9)/(x − 3)

x = 1.50 · h(x) = 4.500

h(x) = (x² − 9)/(x − 3) — a line with a single hole at x = 3
h(x) and your point hole at x = 3 / simplified y = x + 3

\(0/0\) doesn't mean "no limit" — it means the numerator and denominator share a hidden common factor that vanishes at the target point. Remove it algebraically and the limit is usually a plain substitution away.

Why the limit laws hand you back to algebra

In the previous lesson you saw that the quotient limit law only applies when the denominator's limit is nonzero. At a \(0/0\) point, both numerator and denominator head to 0, so the law is simply silent — it doesn't say the limit fails, it says "not my department." And direct substitution is what produced the \(0/0\) in the first place. That's your cue: manipulate the expression first, THEN substitute.

Formal

The key legality: a limit as \(x \to a\) depends only on values of the function for \(x\) near \(a\), never at \(a\) itself. So if two expressions agree at every \(x \neq a\), they have the same limit at \(a\). Cancelling a common factor \((x - a)\) — valid precisely because \(x \neq a\) during the limit process — swaps your undefined-at-\(a\) quotient for an expression that agrees with it everywhere else but is defined at \(a\). In the demo above: \(\dfrac{x^2-9}{x-3} = x + 3\) for every \(x \neq 3\), so both approach the same value, 6, even though only one of them is defined at 3.

Applied

In practice this is pattern recognition. See a \(0/0\) from polynomials? Factor both and cancel the shared factor. See a square-root difference like \(\sqrt{x+4} - 2\)? Multiply top and bottom by the conjugate \(\sqrt{x+4} + 2\) to turn the root difference into plain \(x\)-terms that cancel. See a fraction inside a fraction? Combine the inner fractions over a common denominator first, then cancel. Three forms, three moves — and all three end the same way: substitute into the simplified expression.

The three moves, worked

Factoring you've just seen live in the demo. The conjugate move: for \(\displaystyle\lim_{x \to 0} \frac{\sqrt{x+4}-2}{x}\), multiplying by \(\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\) turns the numerator into \((x+4) - 4 = x\), which cancels the denominator's \(x\), leaving \(\frac{1}{\sqrt{x+4}+2} \to \frac{1}{4}\). The third move — complex fractions — is worked in full below.

Worked example — simplify a complex fraction

Evaluate \(\displaystyle\lim_{x \to 0} \frac{\frac{1}{x+2} - \frac{1}{2}}{x}\). Substitution gives \(\frac{\frac12 - \frac12}{0} = \frac{0}{0}\). Combine the inner fractions over the common denominator \(2(x+2)\): \(\frac{1}{x+2} - \frac{1}{2} = \frac{2 - (x+2)}{2(x+2)} = \frac{-x}{2(x+2)}\). So the whole expression is \(\dfrac{-x}{2(x+2)\,x} = \dfrac{-1}{2(x+2)}\) for \(x \neq 0\), and substituting \(x = 0\) gives \(-\dfrac{1}{4}\).

Your turn — factor and cancel

Evaluate \(\displaystyle\lim_{x \to 5} \frac{x^2 - 25}{x - 5}\). Substitution gives \(\frac{25-25}{5-5} = \frac{0}{0}\), so factor the numerator: \(x^2 - 25 = (x-5)(x+5)\). Cancel the common \((x-5)\), valid for \(x \neq 5\), leaving \(\displaystyle\lim_{x \to 5}\; \) ____ \( = \) ____

Reveal the answer

After cancelling, the expression is \(x + 5\), and \(\displaystyle\lim_{x \to 5} (x+5) = \) 10 — now it's a plain substitution, exactly like sliding along the demo's line to a hole at \((5, 10)\).

More info — why cancelling doesn't "change the function"

It does change the function — at exactly one point. \(\frac{x^2-9}{x-3}\) is undefined at \(x = 3\); \(x + 3\) is defined there. But the limit definition from the start of this section only ever inspects \(x\)-values NEAR the target, never the target itself, so a disagreement confined to the single point \(x = a\) is invisible to the limit. That's also the common pitfall in reverse: don't conclude the original function EQUALS 6 at \(x = 3\) — it has no value there; only its limit is 6. Paul's Online Notes (Dive deeper below) works several more examples of this distinction.

Check your understanding

Question 1 of 4

Evaluate \(\displaystyle\lim_{x \to 4} \frac{x^2 - 16}{x - 4}\).

Question 2 of 4

Evaluate \(\displaystyle\lim_{x \to 0} \frac{\sqrt{x+9} - 3}{x}\).

Question 3 of 4

When you cancel the common factor \((x-a)\) to evaluate \(\lim_{x \to a}\), why is that step legal even though it changes the function at \(x = a\)?

Question 4 of 4

You want \(\displaystyle\lim_{x \to -1} \frac{x^2 - 1}{x + 1}\) and reach for the quotient limit law. What happens?

Recap

  • \(0/0\) from direct substitution is indeterminate, not an answer — simplify algebraically, then substitute again.
  • Polynomials: factor and cancel the common factor (\(\frac{x^2-9}{x-3} = x+3\) for \(x \neq 3\), limit 6).
  • Root differences: multiply by the conjugate (\(\frac{\sqrt{x+4}-2}{x} \to \frac{1}{\sqrt{x+4}+2} \to \frac14\)).
  • Fractions within fractions: combine over a common denominator first, then cancel.
  • Cancelling is legal because the limit ignores the single point \(x = a\) — but the simplified expression may differ from the original AT \(a\).

Dive deeper

Sources

  • Algebraic Techniques for 0/0 Limits