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Teacher Guide: Limits by Direct Substitution

Find limits algebraically: plug in the value when you can, and when you get 0/0, factor and cancel or rationalize first.

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For Teachers

Learning Objectives
  • Evaluate limits of polynomials and rational functions by direct substitution
  • Explain which limit laws justify direct substitution and when the quotient law fails
  • Recognize the indeterminate form 0/0 and tell it apart from nonzero/0
  • Resolve 0/0 by factoring and cancelling a common factor
  • Resolve 0/0 with a square root by multiplying by the conjugate
Prerequisites
  • • Introduction to Limits and Limit Notation
  • • Evaluating Limits Graphically and Evaluating Limits from Tables
  • • Factoring trinomials and differences of squares
  • • Simplifying rational expressions and working with square roots
Discussion Starters
  • 1. Why does the table method from the last lesson and direct substitution give the same number for a polynomial?
  • 2. If substitution gives 00\frac{0}{0}, what does that tell you about the numerator and the denominator as polynomials?
  • 3. Can you invent two different limits that both give 00\frac{0}{0} but have different answers?
  • 4. Why is it allowed to multiply a fraction by x+2x+2\frac{\sqrt{x} + 2}{\sqrt{x} + 2}?
Common Misconceptions

00\frac{0}{0} equals 0, equals 1, or means the limit does not exist

Remediation: Show lim⁡x→02xx=2\lim_{x \to 0} \frac{2x}{x} = 2, lim⁡x→0x2x=0\lim_{x \to 0} \frac{x^2}{x} = 0 and lim⁡x→0xx2\lim_{x \to 0} \frac{x}{x^2} (no finite limit): all three start as 00\frac{0}{0}. Have students check each with a table.

Cancelling (x−a)(x - a) is dividing by zero

Remediation: Stress that the limit never uses x=ax = a. Graph both x2−x−6x−3\frac{x^2 - x - 6}{x - 3} and x+2x + 2: they are the same line except for the hole at x=3x = 3.

Any 0 in the denominator means factor and cancel

Remediation: Give lim⁡x→2x+1x−2\lim_{x \to 2} \frac{x + 1}{x - 2}: substitution gives 30\frac{3}{0}, nothing cancels, and the graph shows a vertical asymptote instead of a hole.

Differentiation Ideas

For Struggling Students:

  • • Start with polynomial limits only, then rational ones with a nonzero denominator
  • • Provide a flowchart: substitute, then number / nonzero over 0 / 0 over 0
  • • Review factoring trinomials with a short warm-up before the 0/0 examples

For On-Level Students:

  • • Mix all three outcomes of substitution in one problem set
  • • Check every answer with the calculator by substituting values like 2.99 and 3.01
  • • Use the graphs to locate the hole whose height equals the limit

For Advanced Students:

  • • Evaluate lim⁡h→0(2+h)2−4h\lim_{h \to 0} \frac{(2 + h)^2 - 4}{h} and connect it to derivatives
  • • Factor sums and differences of cubes, like lim⁡x→2x3−8x−2\lim_{x \to 2} \frac{x^3 - 8}{x - 2}
  • • Rationalize limits with roots in the denominator, like lim⁡x→9x−9x−3\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}
Standards Alignment
  • HSA-SSE.A.2 (CCSS.MATH.CONTENT.HSA.SSE.A.2)

    Use the structure of an expression to identify ways to rewrite it

  • HSA-APR.D.6 (CCSS.MATH.CONTENT.HSA.APR.D.6)

    Rewrite simple rational expressions in different forms

  • HSN-RN.A.2 (CCSS.MATH.CONTENT.HSN.RN.A.2)

    Rewrite expressions involving radicals and rational exponents using the properties of exponents

Lesson Resources
  • visualHole Explorer

    Graphs of (x2−x−6)/(x−3)(x^2 - x - 6)/(x - 3) and (x−2)/(x−4)(\sqrt{x} - 2)/(x - 4) with the hole marked and a table of nearby values

  • activitySubstitute, Sort, Solve

    Students substitute into 12 limit cards and sort them into number, nonzero/0 and 0/0, then solve the 0/0 cards

  • worksheetFactor or Rationalize?

    Mixed 0/0 limits where students first choose the method, then evaluate

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

Direct substitution means finding a limit by simply plugging the value into the function:
lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)
This works for every polynomial and for every rational function whose denominator is not zero at x=ax = a. Their graphs have no gaps there, so the value the function approaches is the value it actually takes.
Why it works: the limit laws. If lim⁡x→af(x)\lim_{x \to a} f(x) and lim⁡x→ag(x)\lim_{x \to a} g(x) exist, then limits can be split up:
  • sum and difference: lim⁡(f±g)=lim⁡f±lim⁡g\lim (f \pm g) = \lim f \pm \lim g
  • constant multiple: lim⁡(c⋅f)=c⋅lim⁡f\lim (c \cdot f) = c \cdot \lim f
  • product: lim⁡(f⋅g)=lim⁡f⋅lim⁡g\lim (f \cdot g) = \lim f \cdot \lim g
  • quotient: lim⁡fg=lim⁡flim⁡g\lim \frac{f}{g} = \frac{\lim f}{\lim g}, only if lim⁡g≠0\lim g \neq 0
  • power: lim⁡fn=(lim⁡f)n\lim f^n = (\lim f)^n
What the substitution tells you:
  • a number, like 55=1\frac{5}{5} = 1: that number is the limit
  • nonzero0\frac{\text{nonzero}}{0}, like 30\frac{3}{0}: no finite limit (the graph has a vertical asymptote, see the lesson Infinite Limits)
  • 00\frac{0}{0}: the indeterminate form 0/0. It is NOT an answer. It tells you to rewrite the expression first.
Fixing 0/0:
  1. 1.Factor and cancel: both parts share a factor (x−a)(x - a). Cancel it, then substitute again.
  2. 2.Rationalize: with a square root like x−2\sqrt{x} - 2, multiply top and bottom by the conjugate x+2\sqrt{x} + 2, cancel, then substitute again.
Cancelling is allowed because the limit only looks at xx values near aa, never at x=ax = a itself, so x−a≠0x - a \neq 0.

Worked Examples

Find lim⁡x→3(x2−2x+3)\lim_{x \to 3} (x^2 - 2x + 3).
1

Check the type of function

x2−2x+3x^2 - 2x + 3 is a polynomial, defined for every xx → Direct substitution works
2

Substitute x=3x = 3

32−2⋅3+3=9−6+33^2 - 2 \cdot 3 + 3 = 9 - 6 + 3 → 66
3

Write the limit

lim⁡x→3(x2−2x+3)=6\lim_{x \to 3} (x^2 - 2x + 3) = 6 → The limit is 66

Common Mistakes

Stopping at 00\frac{0}{0} and answering "the limit is 0", "1" or "does not exist"

Why it's wrong: 00\frac{0}{0} is not a number and not a final answer. It only says that the top and bottom share a factor that hides the real value.

Correct: Treat 00\frac{0}{0} as a signal: factor and cancel (or rationalize), then substitute again. lim⁡x→3x2−x−6x−3=5\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3} = 5.

Cancelling terms instead of factors, like x2−9x−3=x−9−3\frac{x^2 - 9}{x - 3} = \frac{x - 9}{-3}

Why it's wrong: You may only cancel a factor that multiplies the whole numerator and the whole denominator. xx inside a sum is not such a factor.

Correct: Factor first: x2−9x−3=(x−3)(x+3)x−3=x+3\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3, so the limit as x→3x \to 3 is 66.

Treating 30\frac{3}{0} like 00\frac{0}{0} and trying to factor

Why it's wrong: When only the denominator is 0, there is no common factor to cancel. The values grow without bound near that xx.

Correct: nonzero0\frac{\text{nonzero}}{0} means there is no finite limit (vertical asymptote). Only 00\frac{0}{0} calls for factoring or rationalizing.

Multiplying only the numerator by the conjugate

Why it's wrong: Multiplying just one part changes the value of the fraction. You are allowed to multiply by x+2x+2\frac{\sqrt{x} + 2}{\sqrt{x} + 2}, which equals 1.

Correct: Multiply top AND bottom by the conjugate, and leave the bottom factored so you can cancel (x−4)(x - 4).

Why It Matters

In the lessons Evaluating Limits Graphically and Evaluating Limits from Tables you estimated limits by looking. Direct substitution gives the exact value in seconds, and it is the method you will use most often in calculus.
  • Derivatives: every derivative starts as a limit of the form 00\frac{0}{0}, for example lim⁡h→0(3+h)2−9h\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}. Factoring and cancelling is exactly how you solve it.
  • Physics: instantaneous speed is the limit of average speeds, found by factoring a difference quotient.
  • Continuity: a function is continuous at aa exactly when direct substitution works there (lesson Continuity).
  • Holes in graphs: the factor you cancel shows where a rational function has a hole, and the limit tells you its height.
Knowing when you may substitute, and what to do when you get 00\frac{0}{0}, saves you from wrong answers like "the limit is 0" or "the limit does not exist".

Real World Applications

Instantaneous Speed

A speedometer shows speed at one instant. Physicists find it as the limit of average speeds over shorter and shorter time intervals, and that limit always starts as 0/0.

Example:

A cart moves so that its position is s(t)=t2+3ts(t) = t^2 + 3t meters after tt seconds. Its average speed between second 2 and time tt is s(t)−s(2)t−2=t2+3t−10t−2\frac{s(t) - s(2)}{t - 2} = \frac{t^2 + 3t - 10}{t - 2}. Factoring gives (t−2)(t+5)t−2=t+5\frac{(t - 2)(t + 5)}{t - 2} = t + 5.

1Try It Yourself

The cart's position is s(t)=t2+3ts(t) = t^2 + 3t meters. Its speed at t=2t = 2 seconds is lim⁡t→2t2+3t−10t−2\lim_{t \to 2} \frac{t^2 + 3t - 10}{t - 2}.

What is the cart's speed at t=2t = 2 seconds, in meters per second?

Step 1: Write the mathematical expression

Factor, cancel and substitute: the speed is

Growing Square Tiles

Designers and engineers ask how fast an area changes when a length changes. For a square of side $x$ the area is $x^2$, and the rate of change at one size is a 0/0 limit.

Example:

A square tile with side xx cm has area A(x)=x2A(x) = x^2 cm². When the side grows from 5 cm to xx cm, the area changes on average by x2−25x−5\frac{x^2 - 25}{x - 5} cm² per cm of side.

2Try It Yourself

The rate at which the tile's area grows when the side is exactly 5 cm is lim⁡x→5x2−25x−5\lim_{x \to 5} \frac{x^2 - 25}{x - 5}.

How many cm² per cm does the area grow at a side of 5 cm?

Step 1: Write the mathematical expression

Factor x2−25x^2 - 25, cancel and substitute: the rate is

Plant Growth Rate

Biologists model growth with square root formulas. The growth rate on a given day is a limit that needs rationalizing.

Example:

A seedling is h(t)=th(t) = \sqrt{t} dm tall after tt days. Its average growth between day 9 and day tt is t−3t−9\frac{\sqrt{t} - 3}{t - 9} dm per day.

3Try It Yourself

The growth rate on day 9 is lim⁡t→9t−3t−9\lim_{t \to 9} \frac{\sqrt{t} - 3}{t - 9}. Multiplying top and bottom by t+3\sqrt{t} + 3 turns it into lim⁡t→91t+3\lim_{t \to 9} \frac{1}{\sqrt{t} + 3}.

What is the growth rate on day 9, in dm per day?

Step 1: Write the mathematical expression

Substitute into the simplified expression: the rate is

Key Takeaways

  • 1For polynomials, lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a): just substitute
  • 2For rational functions, substitute when the denominator is not 0 at x=ax = a
  • 3The limit laws let you split a limit into sums, products, quotients (denominator limit not 0) and powers
  • 400\frac{0}{0} is an indeterminate form, never the answer: rewrite the expression first
  • 5Factor and cancel the common factor (x−a)(x - a), or multiply by the conjugate when there is a square root, then substitute again
  • 6nonzero0\frac{\text{nonzero}}{0} means no finite limit (vertical asymptote), not a hole

Frequently Asked Questions

Why may I cancel (x−3)(x - 3) if it is 0 at x=3x = 3?

A limit as x→3x \to 3 only uses xx values close to 3, never x=3x = 3 itself. For all those values x−3≠0x - 3 \neq 0, so cancelling is a legal step. The two expressions agree everywhere except at x=3x = 3, and the limit does not care about that single point.

If I get 00\frac{0}{0}, does the limit exist?

You cannot tell yet. 00\frac{0}{0} can hide any value: lim⁡x→02xx=2\lim_{x \to 0} \frac{2x}{x} = 2, lim⁡x→0x2x=0\lim_{x \to 0} \frac{x^2}{x} = 0. Simplify first, then decide.

When should I rationalize instead of factoring?

When the 0 comes from a difference with a square root, like x−2\sqrt{x} - 2 or x+4−2\sqrt{x + 4} - 2. Multiply top and bottom by the conjugate (same terms, opposite sign in the middle), because (x−2)(x+2)=x−4(\sqrt{x} - 2)(\sqrt{x} + 2) = x - 4 has no root left.

Is the limit the same as the function value?

For polynomials and for rational functions where the denominator is not 0, yes: that is exactly what direct substitution says. At a hole, the function value does not exist, but the limit does (lessons Continuity and Types of Discontinuity go deeper).

Glossary

Direct substitution
Finding lim⁡x→af(x)\lim_{x \to a} f(x) by computing f(a)f(a); works for polynomials and for rational functions with a nonzero denominator at aa
Limit laws
Rules that split a limit of a sum, difference, product, quotient or power into limits of its parts
Indeterminate form 0/0
The result 00\frac{0}{0} after substitution; it does not give a value and means the expression must be rewritten
Conjugate
For x−b\sqrt{x} - b the conjugate is x+b\sqrt{x} + b; their product x−b2x - b^2 has no square root
Rationalizing
Multiplying top and bottom by a conjugate to remove a square root from a difference

Formula Card

Direct substitution

lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)

For polynomials, and for rational functions when the denominator is not 0 at $a$

Quotient law

lim⁡x→af(x)g(x)=lim⁡f(x)lim⁡g(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}

Only when $\lim_{x \to a} g(x) \neq 0$

Factor and cancel

(x−a)⋅p(x)(x−a)⋅q(x)=p(x)q(x)\frac{(x - a) \cdot p(x)}{(x - a) \cdot q(x)} = \frac{p(x)}{q(x)}

Use when substitution gives $\frac{0}{0}$, then substitute again

Conjugate product

(x−b)(x+b)=x−b2(\sqrt{x} - b)(\sqrt{x} + b) = x - b^2

Use to rationalize a 0/0 limit with a square root

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Evaluating Limits from Tables

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