Unit 1 · Topic 1.6

1.6 Determining Limits Using Algebraic Manipulation

When direct substitution gives 0/0, the limit is not decided yet: 0/0 is called an indeterminate form. You rewrite the expression into an equivalent one that agrees with it everywhere except x = c, then substitute. Factoring, conjugates, combining fractions and trig identities are the main tools.

Key terms

indeterminate form 0/0
factoring and canceling
conjugate
rationalizing
trig identities

Why 0/0 isn't an answer

If the top and bottom both approach 0, the quotient could approach anything: 1, 5, 0, ∞ or nothing at all. So 0/0 means “more work needed.” It is not 0, not 1, and not “does not exist.”

The fix is to find a simpler expression that equals the original one for every x near c except c itself. Since limits ignore x = c, the two expressions have the same limit.

After rewriting, substitute again. If you get a real number, you're done. If you still get 0/0, there may be another common factor, so rewrite again. If you get (nonzero)/0, the function is unbounded near c, so look at signs on each side (1.14).

Tool 1: factor and cancel

If a polynomial is 0 at x = c, then (x − c) is a factor. Factor the top and bottom, cancel the common (x − c), and substitute.

Useful patterns: a² − b² = (a − b)(a + b), a³ − b³ = (a − b)(a² + ab + b²) and a³ + b³ = (a + b)(a² − ab + b²).

Tool 2: multiply by the conjugate

For square roots like √(x + 9) − 3, multiply top and bottom by the conjugate, √(x + 9) + 3. The product (√A − B)(√A + B) = A − B² removes the root and usually exposes a factor that cancels. This is called rationalizing.

Tool 3: combine fractions

For a complex fraction (a fraction inside a fraction), get a common denominator in the top, simplify, then cancel. For example, 1/(x + 4) − 1/4 = (4 − (x + 4)) / (4(x + 4)) = −x / (4(x + 4)).

Tool 4: trig identities

Pythagorean identities often turn a trig 0/0 into something that cancels. For instance, sin²x = 1 − cos²x = (1 − cos x)(1 + cos x). So sin²x / (1 − cos x) = 1 + cos x when cos x ≠ 1, and the limit as x→0 is 2.

Two special limits are also worth knowing: lim (x→0) sin x / x = 1 and lim (x→0) (1 − cos x) / x = 0. Topic 1.8 shows where the first one comes from.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Factor and cancel
Find lim (x→3) (x² − x − 6) / (x² − 9).
Show the solutionHide the solution
1. Step 1: Direct substitution: (9 − 3 − 6) / (9 − 9) = 0/0, so rewrite.
2. Step 2: Factor: x² − x − 6 = (x − 3)(x + 2) and x² − 9 = (x − 3)(x + 3).
3. Step 3: For x ≠ 3, the expression equals (x + 2) / (x + 3).
4. Step 4: Substitute x = 3: 5/6.
Answer: lim (x→3) (x² − x − 6) / (x² − 9) = 5/6
Example 2
Trap: you can't cancel inside a square root
Find lim (x→0) (√(x + 9) − 3) / x.
Show the solutionHide the solution
1. Step 1: A tempting shortcut is to split the root: √(x + 9) − 3 = √x + 3 − 3 = √x. That is false, because √(x + 9) is not √x + 3. (Try x = 16: √25 = 5, but √16 + 3 = 7.)
2. Step 2: Direct substitution gives (3 − 3) / 0 = 0/0, so you need a real rewrite.
3. Step 3: Multiply top and bottom by the conjugate, √(x + 9) + 3. The top becomes (x + 9) − 9 = x.
4. Step 4: So the expression is x / (x(√(x + 9) + 3)) = 1 / (√(x + 9) + 3) for x ≠ 0.
5. Step 5: Substitute x = 0: 1 / (3 + 3) = 1/6.
Answer: lim (x→0) (√(x + 9) − 3) / x = 1/6
Example 3
Combine fractions
Find lim (x→0) [1/(x + 4) − 1/4] / x.
Show the solutionHide the solution
1. Step 1: Direct substitution: (1/4 − 1/4) / 0 = 0/0.
2. Step 2: Combine the top over 4(x + 4): [4 − (x + 4)] / (4(x + 4)) = −x / (4(x + 4)).
3. Step 3: Divide by x: −x / (4x(x + 4)) = −1 / (4(x + 4)) for x ≠ 0.
4. Step 4: Substitute x = 0: −1/16.
Answer: lim (x→0) [1/(x + 4) − 1/4] / x = −1/16

Common mistakes

Writing “0/0, so the limit does not exist.” 0/0 means the limit is undecided, not missing.
Dropping the “lim” in the middle of the work, like writing (x − 3)(x + 2)/(x − 3)(x + 3) = 5/6. Keep writing lim (x→3) until you actually substitute.
Multiplying only the top by the conjugate. You must multiply top and bottom by the same thing.

On the exam

These limits are a staple of the no-calculator multiple-choice section. A quick check: if substitution gives 0/0, look for a factor of (x − c).
Limits in this form often turn out to be a derivative in disguise (2.2). Unit 4 also gives L'Hospital's Rule as another tool for 0/0.

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