Grade by Grade
A. Rate of Change · Limits and Rates of Change

Indeterminate Forms: Rationalizing and Other Strategies

A1.4About 70 minutes
Learning goals
  • I can evaluate 0/0 limits that contain square roots by multiplying by a conjugate.
  • I can evaluate limits with absolute values by finding one-sided limits.
  • I can use a change of variable to evaluate limits with cube roots and other roots.
  • I can choose an appropriate strategy for an indeterminate limit.

Words to know

conjugate
For √A − B, the conjugate is √A + B. Their product is A − B², which has no square root.
rationalize
Rewrite an expression to remove a square root from a numerator or denominator, usually by multiplying by a conjugate.
absolute value
|x| = x if x ≥ 0 and |x| = −x if x < 0. It is a piecewise function.
change of variable
Replacing an expression such as ∛x by a new letter u to make the algebra simpler.
one-sided limit
A limit as x approaches a from only the left (a⁻) or only the right (a⁺).

When factoring will not do it

Try lim (x → 4) (√x − 2)/(x − 4). Substituting gives 0/0. But √x − 2 is not a polynomial, so there is nothing obvious to factor. We need a new tool.

The trick comes from the difference of squares: (√x − 2)(√x + 2) = (√x)² − 2² = x − 4. Multiplying by √x + 2 turns the square root into exactly the factor we want to cancel. The expression √x + 2 is called the conjugate of √x − 2.

Remember from Grade 11: in the MCR3U unit Trigonometric Ratios, Identities and Laws you rationalized denominators such as 1/√2 = √2/2, and in Functions and Equivalent Expressions you simplified rational expressions and stated restrictions. Here you will rationalize numerators too.

Worked example

Example 1: Rationalizing the numerator

Evaluate lim (x → 4) (√x − 2)/(x − 4).

  1. Substitute: 0/0.
  2. Multiply the top and bottom by the conjugate √x + 2: (√x − 2)(√x + 2)/((x − 4)(√x + 2)).
  3. The top becomes x − 4: (x − 4)/((x − 4)(√x + 2)).
  4. Cancel x − 4 (x ≠ 4): 1/(√x + 2).
  5. Substitute: 1/(2 + 2) = 1/4.

1/4. This is also the slope of the tangent to y = √x at (4, 2), which you estimated in earlier lessons.

0246810120.10.20.30.40.5xyhole at (4, 1/4)y = (√x − 2)/(x − 4)Multiply by the conjugate:(√x − 2)(√x + 2) = x − 4so y = 1/(√x + 2), x ≠ 4lim y = 1/(2 + 2) = 1/4x→4
The graph of y = (√x − 2)/(x − 4) has a hole at (4, 1/4). Multiplying by the conjugate shows it equals 1/(√x + 2) everywhere else.
Worked example

Example 2: A root with a constant inside

Evaluate lim (x → 0) (√(x + 9) − 3)/x.

  1. Substitute: (3 − 3)/0 = 0/0.
  2. Multiply by the conjugate √(x + 9) + 3. The top becomes (x + 9) − 9 = x.
  3. So the expression is x/(x(√(x + 9) + 3)) = 1/(√(x + 9) + 3) for x ≠ 0.
  4. Substitute: 1/(3 + 3) = 1/6.

1/6

Worked example

Example 3: Absolute values

Evaluate lim (x → 2) (x² − 4)/|x − 2|, or explain why it does not exist.

  1. |x − 2| has different formulas on each side of 2, so find one-sided limits.
  2. Right side (x > 2): |x − 2| = x − 2. The expression is (x − 2)(x + 2)/(x − 2) = x + 2 → 4.
  3. Left side (x < 2): |x − 2| = −(x − 2). The expression is (x − 2)(x + 2)/(−(x − 2)) = −(x + 2) → −4.
  4. The one-sided limits are different.

lim (x → 2⁺) = 4 and lim (x → 2⁻) = −4, so the limit does not exist.

Worked example

Example 4: A change of variable

Evaluate lim (x → 1) (∛x − 1)/(x − 1).

  1. Substitute: 0/0. A conjugate for a cube root is messy, so change the variable.
  2. Let u = ∛x. Then x = u³, and as x → 1, u → 1.
  3. The expression becomes (u − 1)/(u³ − 1) = (u − 1)/((u − 1)(u² + u + 1)) = 1/(u² + u + 1) for u ≠ 1.
  4. Substitute u = 1: 1/(1 + 1 + 1) = 1/3.

1/3

FactorUse for polynomials,x² − a², x³ − a³ExpandUse for (a + h)ⁿ in adifference quotientCommon denominatorUse for fractions insidea fractionConjugateUse for square roots:multiply by √A + BOne-sided limitsUse for absolute values,piecewise rulesSubstitute uUse for cube or fourthroots: u = ∛x …
Six strategies for 0/0 limits and the clues that tell you which one to try. After any of them, substitute again.

Choosing a strategy

Look at what is in the expression:

  • Only polynomials → factor (or expand, if there is a (a + h)ⁿ).
  • Fractions inside a fraction → common denominator.
  • A square root next to a + or − sign → multiply by the conjugate.
  • Absolute value or a piecewise rule → one-sided limits.
  • Cube roots or fourth roots → change of variable.

Sometimes you need two strategies in a row. If you are stuck, a quick table of values with x = a ± 0.001 tells you what answer to aim for.

Just for fun

Multiplying by the conjugate is the algebra version of bringing a friend to a party just so they can cancel you out of the awkward conversation. √x + 2 shows up, and suddenly x − 4 can leave.

Hands-on activity

Check algebra with a calculator

You need: a scientific calculator and paper.

  1. Choose any three limits from this lesson's examples or practice.
  2. For each one, evaluate the original expression at x = a + 0.001 and x = a − 0.001 (for h-limits, use h = ±0.001). Write down at least six decimal places.
  3. Compare with your exact algebraic answer, written as a decimal.
  4. Now try x = a + 0.000 000 000 001 on your calculator for one of them. What happens?

What to look for: with ±0.001, the values should agree with the exact answer to about three or more decimal places (for example 0.249 98 and 0.250 02 for an answer of 1/4). With an extremely small step, the calculator may return 0 or a wildly wrong value, because it runs out of digits when subtracting nearly equal numbers. That is why the algebra matters: it gives the exact value with no rounding at all.

Your turn

1

Evaluate lim (x → 9) (√x − 3)/(x − 9).

2

Evaluate lim (x → 0) (√(x + 4) − 2)/x.

3

Evaluate lim (x → 1) (x − 1)/(√x − 1).

4

Evaluate lim (h → 0) (√(16 + h) − 4)/h.

5

Find lim (x → 5⁺) |x − 5|/(x − 5) and lim (x → 5⁻) |x − 5|/(x − 5).

6

Evaluate lim (x → 8) (∛x − 2)/(x − 8).

7

Evaluate lim (x → 0) (2 − √(4 − x))/x.

More practice: warm-up

1

Write the conjugate of √x − 5.

2

Expand (√x − 3)(√x + 3).

3

Expand (√(x + 1) − 1)(√(x + 1) + 1).

4

Find lim (x → 0⁺) |x|/x.

5

Find lim (x → 4) (√x + 2).

More practice: on your own

1

Evaluate lim (x → 3) (√(x + 1) − 2)/(x − 3).

2

Evaluate lim (x → 2) (x² − 4)/(√(x + 2) − 2).

3

Evaluate lim (x → 0) (√(1 + x) − √(1 − x))/x.

4

Evaluate lim (x → −2) (x + 2)/|x + 2|, or explain why it does not exist.

5

Evaluate lim (x → 1) (x^(1/4) − 1)/(x − 1).

Challenge

1

Evaluate lim (x → ∞) (√(x² + x) − x).

2

Evaluate lim (x → 0) (√(x + 1) − 1)/(∛(x + 1) − 1).

3

Evaluate lim (h → 0) [1/√(4 + h) − 1/2]/h.

Check your understanding

Quiz

1.

What is the conjugate of √(x + 5) − 3?

2.

Evaluate lim (x → 25) (√x − 5)/(x − 25).

3.

Evaluate lim (x → 0) (√(x + 1) − 1)/x.

4.

What is lim (x → 3⁻) |x − 3|/(x − 3)?

5.

Which strategy fits lim (x → 1) (∛x − 1)/(x − 1) best?

6.

Evaluate lim (h → 0) (√(9 + h) − 3)/h.

For parents and teachers

Big idea (A1.4): when a square root causes 0/0, multiplying by the conjugate removes it; absolute values need one-sided limits; other roots yield to a change of variable. These complete the toolkit for evaluating limits algebraically.

How to help: Ask your teen to say which strategy they will use and why before starting. Then check the final answer numerically with x = a ± 0.001, as in the activity.

Common difficulties:

  • Multiplying only the numerator by the conjugate (it must be top and bottom, so you multiply by 1).
  • Expanding the conjugate product fully when only the difference of squares is needed.
  • Writing √(x + 9) − 3 as √x + 3 − 3; roots do not split over addition.
  • Forgetting that |x − a| = −(x − a) on the left side.