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

Indeterminate Forms: Factoring and Simplifying

A1.4 A1.6About 70 minutes
Learning goals
  • I can recognize the indeterminate form 0/0 and explain why it needs more work.
  • I can evaluate limits by factoring and cancelling a common factor.
  • I can evaluate limits by expanding and simplifying, including difference quotients.
  • I can evaluate limits of compound fractions by finding a common denominator.

Words to know

indeterminate form
An expression like 0/0 that a limit can produce on substitution, which does not determine the limit's value.
common factor
A factor, such as (x − a), shared by the numerator and denominator. Cancelling it is valid when x ≠ a.
difference of squares
a² − b² = (a − b)(a + b).
sum and difference of cubes
a³ − b³ = (a − b)(a² + ab + b²) and a³ + b³ = (a + b)(a² − ab + b²).
factor theorem
x − a is a factor of a polynomial p(x) exactly when p(a) = 0.
compound fraction
A fraction whose numerator or denominator contains fractions.

The most interesting 0/0

Try substituting x = 3 into (x² − 9)/(x − 3). You get 0/0. Is the limit 0? 1? Undefined?

Look at three limits that all give 0/0 when you substitute x = 0:

  • x/x → 1
  • 2x/x → 2
  • x²/x = x → 0

Same "0/0", three different answers. That is why 0/0 is called an indeterminate form: it tells you the limit might exist, but not what it is. You must rewrite the expression first.

This matters a lot: every instantaneous rate of change, (f(a + h) − f(a))/h as h → 0, gives 0/0 when you substitute h = 0. Calculus lives inside 0/0.

Why cancelling is allowed

In lim (x → 3) (x − 3)(x + 3)/(x − 3), x approaches 3 but is never equal to 3. So x − 3 is never 0, and we may cancel it:

(x − 3)(x + 3)/(x − 3) = x + 3, for x ≠ 3.

The two functions agree everywhere except at x = 3. Limits only care about values near 3, so the limits are equal, and lim (x → 3) (x + 3) = 6 by substitution.

01234502468xyhole at (3, 6)y = (x² − 9)/(x − 3)= (x − 3)(x + 3)/(x − 3)= x + 3, for x ≠ 3Same graph as y = x + 3,except for one missing point.lim (x² − 9)/(x − 3) = 6x→3
The graph of (x² − 9)/(x − 3) is the line y = x + 3 with a hole at (3, 6). The limit is the height of the hole.
Worked example

Example 1: Factor and cancel

Evaluate lim (x → −1) (x² + 3x + 2)/(x² − 1).

  1. Substitute: (1 − 3 + 2)/(1 − 1) = 0/0. Indeterminate, so rework.
  2. Factor: (x + 1)(x + 2)/((x + 1)(x − 1)).
  3. Cancel x + 1 (allowed because x ≠ −1): (x + 2)/(x − 1).
  4. Substitute: (−1 + 2)/(−1 − 1) = 1/(−2).

−1/2

Worked example

Example 2: Difference of cubes

Evaluate lim (x → 2) (x³ − 8)/(x² − 4).

  1. Substitute: 0/0.
  2. x³ − 8 = (x − 2)(x² + 2x + 4) and x² − 4 = (x − 2)(x + 2).
  3. Cancel x − 2: (x² + 2x + 4)/(x + 2).
  4. Substitute: (4 + 4 + 4)/4 = 12/4 = 3.

3

When a factor is hard to see

If p(a) = 0 for a polynomial p, the factor theorem (from the MHF4U lesson The Factor Theorem: Factoring Cubics and Quartics in Polynomial Functions and Equations) guarantees that x − a is a factor. Divide by x − a with long division or synthetic division to find the other factor. Example: x⁴ − 1 at x = 1 is 0, and x⁴ − 1 = (x − 1)(x³ + x² + x + 1).

Worked example

Example 3: Expand and simplify (a difference quotient)

Evaluate lim (h → 0) ((2 + h)³ − 8)/h. What does it represent?

  1. Substitute h = 0: (8 − 8)/0 = 0/0.
  2. Expand: (2 + h)³ = 8 + 12h + 6h² + h³.
  3. Numerator: 8 + 12h + 6h² + h³ − 8 = 12h + 6h² + h³ = h(12 + 6h + h²).
  4. Cancel h (h ≠ 0): 12 + 6h + h².
  5. Substitute h = 0: 12.

12. This is the difference quotient of f(x) = x³ at a = 2, so the tangent to y = x³ at (2, 8) has slope 12. The lesson The Difference Quotient as a Limit builds on this.

Worked example

Example 4: A compound fraction

Evaluate lim (x → 0) [1/(x + 2) − 1/2] / x.

  1. Substitute: (1/2 − 1/2)/0 = 0/0.
  2. Combine the top over the common denominator 2(x + 2): 1/(x + 2) − 1/2 = (2 − (x + 2))/(2(x + 2)) = −x/(2(x + 2)).
  3. Divide by x: −x/(2(x + 2)) × 1/x = −1/(2(x + 2)) for x ≠ 0.
  4. Substitute x = 0: −1/(2 × 2) = −1/4.

−1/4

Substitute x = aa real numberthat is the limit(non-zero) ÷ 0check each side: ±∞0 ÷ 0indeterminate: reworkfactor and cancel,expand, combine fractions…substitute again (usually gives a number)
A plan for any limit: substitute first. A number is the answer; (non-zero)/0 means an asymptote; 0/0 means rework the expression and substitute again.

Watch out: 0/0 is not always rescued

After cancelling, substitute again. If you now get (non-zero)/0, the limit does not exist as a number. Example: lim (x → 2) (x² − 4)/(x − 2)² = lim (x + 2)/(x − 2), which gives 4/0. Check the sides: the limit is ∞ from the right and −∞ from the left.

Just for fun

0/0 is like a box labelled "could be anything". Factoring is the box cutter. And, as with any box, cut away from yourself: forgetting the x ≠ a condition is how algebra gets hurt.

Hands-on activity

Shrinking rectangles

You need: paper, a ruler, a pencil, and scissors (use them carefully).

The expression x² + 3x − 10 factors as (x − 2)(x + 5). Think of it as the area of a rectangle with width x − 2 and length x + 5 (in centimetres).

  1. For x = 4, draw and cut out the rectangle: width 2 cm, length 9 cm. Check that the area is 4² + 3(4) − 10 = 18 cm².
  2. Repeat for x = 3 (width 1 cm, length 8 cm, area 8 cm²) and x = 2.5 (width 0.5 cm, length 7.5 cm, area 3.75 cm²).
  3. For each rectangle, calculate area ÷ width. Compare it with the length.
  4. Imagine x getting closer and closer to 2. What happens to the width? To the area? To area ÷ width?
  5. Write what you found as a limit.

What to look for: area ÷ width always equals the length, x + 5 (9, 8 and 7.5 cm). As x → 2, the width and the area both shrink toward 0, but area ÷ width approaches 2 + 5 = 7. So lim (x → 2) (x² + 3x − 10)/(x − 2) = 7. At exactly x = 2 there is no rectangle at all: that is the 0/0, and the limit is the length the strips approach.

Your turn

1

Evaluate lim (x → 5) (x² − 25)/(x − 5).

2

Evaluate lim (x → −2) (x² + 5x + 6)/(x + 2).

3

Evaluate lim (x → 1) (x² + x − 2)/(x² − 1).

4

Evaluate lim (x → 3) (x³ − 27)/(x − 3).

5

Evaluate lim (h → 0) ((4 + h)² − 16)/h.

6

Evaluate lim (x → 0) [1/(x + 3) − 1/3]/x.

7

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

More practice: warm-up

1

What do you get when you substitute x = 2 into (x² − 4)/(x − 2)?

2

Factor x² − 16.

3

Evaluate lim (x → 4) (x² − 16)/(x − 4).

4

Evaluate lim (x → 0) (x² + 3x)/x.

5

Evaluate lim (x → 1) (x − 1)/(x² − 1).

More practice: on your own

1

Evaluate lim (x → 1/2) (2x² + x − 1)/(2x − 1).

2

Evaluate lim (x → 1) (x⁴ − 1)/(x − 1).

3

Evaluate lim (h → 0) ((1 + h)³ − 1)/h.

4

Evaluate lim (x → −3) (x³ + 27)/(x + 3).

5

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

Challenge

1

Evaluate lim (x → 2) (x³ − 3x² + 4)/(x² − 4x + 4).

2

Use xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + axⁿ⁻² + … + aⁿ⁻¹) to show lim (x → a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹.

3

Find b and c if lim (x → 3) (x² + bx + c)/(x − 3) = 7.

Check your understanding

Quiz

1.

Substituting into a limit gives 0/0. What can you conclude?

2.

Evaluate lim (x → 6) (x² − 36)/(x − 6).

3.

Evaluate lim (x → 1) (x³ − 1)/(x − 1).

4.

Evaluate lim (h → 0) ((3 + h)² − 9)/h.

5.

Why can we cancel x − a when finding lim (x → a)?

6.

Evaluate lim (x → 0) [1/(x + 4) − 1/4]/x.

For parents and teachers

Big idea (A1.4, A1.6): 0/0 does not mean the limit fails; it means the expression must be rewritten. Factoring and cancelling a common factor, expanding a difference quotient, or combining fractions usually removes the 0/0. These are the algebraic skills behind every instantaneous rate of change.

How to help: Have your teen explain, out loud, why cancelling x − 3 is allowed in a limit but would be wrong if x could equal 3. Then check one answer with a calculator by substituting x = a ± 0.001.

Common difficulties:

  • Writing "0/0 = 0" or "0/0 = 1".
  • Factoring errors, especially with cubes. Expand the factored form to check.
  • Errors in expanding (a + h)³; use (a + h)³ = a³ + 3a²h + 3ah² + h³.
  • Sign slips when combining fractions such as 1/(x + 2) − 1/2.