Replaces numerical estimation with exact computation. Gives the properties that let a limit pass through sums, products, quotients and powers, establishes substitution as the first move, and develops the algebraic techniques for resolving indeterminate forms.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 12 — Introduction to Calculus
§12.2 Finding Limits: Properties of Limits, pp. 1404-1415
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1404-1415 — the pages these objectives are drawn from
Warm-up
The previous section built tables. That was a lot of work.
Discussion prompt
To find the limit of a polynomial at a point, why might substitution be enough?
Hint: What does the graph do there?
Answer:
A polynomial has no holes or jumps, so the outputs near a point close in on the output at the point.
So the limit and the value agree, and substituting gives the limit directly — no table needed.
That works for most functions at most points. The interesting cases are exactly where it fails, and this section is about recognising and handling those.
Concept
Most limits are found by substituting the point. When substitution produces an indeterminate form instead, that form indicates which algebraic technique will resolve it.
\[ \frac{0}{0}: \text{ indeterminate} \quad\text{versus}\quad \frac{5}{0}: \text{ undefined} \]
The distinction between those two is the section's central point. One says the limit may well exist and needs work; the other says the outputs grow without bound and no work will help.
Figure (svg): A diagram showing substitution working directly for most limits and producing an indeterminate form for the interesting ones
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1404-1407
Section
Section 1
Concept
The limit of a sum, product, quotient or power is obtained by taking limits of the parts, subject to one condition on quotients.
The quotient condition is the only one that ever bites in practice, and it bites exactly when substitution gives zero on the bottom. The rest of the section is about what to do then.
Figure (svg): Four cards giving the limit properties for sums, products, quotients and powers
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1404-1408
Picture it
Each lets the limit move past an operation.
Figure (svg): Four cards giving the limit properties for sums, products, quotients and powers
The caption identifies the one that carries a condition. Quotients are where substitution fails, and the whole rest of the section addresses that case.
Worked example
Break the expression into parts.
\[ \text{Evaluate } \lim_{x\to 3}(x^2+2x-1). \]
Split by the sum property
Why: Three separate limits.
Evaluate each
Why: By substitution.
\[ 9, 6, -1 \]
Recombine
Why: By the same property.
\[ 9 + 6 - 1 \]
Compute
Why: The total.
\[ 14 \]
Figure (svg): Four cards giving the limit properties for sums, products, quotients and powers
\[ 14 \]
Verify: check by substituting directly
Why: Substituting three into the whole expression gives nine plus six minus one, which is 14 — the same answer with less writing. The properties justify the shortcut rather than replacing it.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1405-1407
Prediction
You want the limit of a quotient.
Predict first
What must be true?
Correct: The denominator's limit must be nonzero.
Why: Dividing by zero is undefined, so the property carries that condition. When it fails, the numerator's limit then decides between an indeterminate form and a genuine failure.
Worked example
The property has a precondition.
\[ \text{When does the quotient property fail?} \]
State the property
Why: Limit of a quotient.
Note the condition
Why: The bottom limit must be nonzero.
Consider what happens otherwise
Why: Division by zero.
Conclude
Why: The property does not apply.
Figure (svg): The solution to Worked example the quotient condition shown as a ladder of expressions, one row per legal move
\[ \text{fails when the bottom limit is }0 \]
Verify: distinguish the two cases
Why: If the top limit is nonzero the outputs grow without bound and no limit exists. If it is also zero the form is indeterminate and algebra may still find the limit — two quite different situations behind the same failed property.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1407-1408
Trap
\[ \lim\frac{f}{g}=\frac{\lim f}{\lim g}, \text{ always} \]
Use the property unconditionally
Why: The denominator's limit is not checked.
When that limit is zero the property does not apply at all.
Check the denominator's limit first. The property requires it to be nonzero.
If it is zero, look at the numerator: a nonzero top means no limit, and a zero top means an indeterminate form to resolve.
That check is the branch point for everything else in the section, which is why it is worth making explicitly.
Sorting
Only one of the four does.
Sort into buckets
Sort each property.
Faded example
Splitting a polynomial limit.
Fill in the blanks
\lim(x^2+2x-1)=9+6-1=14
Why: The sum property lets each term's limit be taken separately and the results recombined. For a polynomial each part is found by substitution, which is why the whole thing can be substituted at once.
Step zero
You are asked to evaluate a limit.
Discussion prompt
What do you try before anything else?
Hint: The simplest thing.
Answer:
Substitute the point. For most functions at most points that gives the limit directly.
If it produces a number, that is the answer. If it produces zero over zero, that is a signal to do algebra.
And if it produces a nonzero number over zero, the limit does not exist. All three outcomes are informative, which is why substitution is worth trying first even when it will fail.
Section
Section 2
Concept
Substituting the point gives the limit whenever the function's value and limit agree there, which is the case for polynomials everywhere and for most functions at most points.
Naming the property that makes substitution valid — continuity — is what §12.3 does. Until then it is enough to know that substitution works for the standard functions except at the points where they misbehave.
Figure (svg): A diagram showing substitution working directly for most limits and producing an indeterminate form for the interesting ones
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1408-1411
Picture it
Two outcomes, both informative.
Figure (svg): A diagram showing substitution working directly for most limits and producing an indeterminate form for the interesting ones
The bottom line is the section's organising idea. An indeterminate form is a signal, not an answer, and it points to the technique that will resolve it.
Worked example
The denominator does not vanish.
\[ \text{Evaluate } \lim_{x\to 2}\frac{x^2+1}{x-1}. \]
Check the denominator
Why: At two.
\[ 1,\text{ nonzero} \]
Apply the quotient property
Why: The condition holds.
Substitute
Why: Both parts.
\[ 5\text{ over } 1 \]
Compute
Why: The limit.
\[ 5 \]
Figure (svg): A diagram showing substitution working directly for most limits and producing an indeterminate form for the interesting ones
\[ 5 \]
Verify: confirm the condition was checked
Why: The denominator's limit is 1, which is nonzero — so the quotient property applied and substitution was legitimate. Checking that before substituting is what distinguishes a valid computation from a lucky one.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1409-1410
Prediction
Substitution gives zero over zero.
Predict first
What does that tell you?
Correct: The limit may exist; algebra is needed to find it.
Why: Zero over zero determines nothing — two expressions producing it can have different limits. It signals that the numerator and denominator share a vanishing factor, which algebra can remove.
Worked example
The form says what to do.
\[ \text{What does substituting give for } \lim_{x\to 2}\frac{x^2-4}{x-2}? \]
Substitute into the numerator
Why: Four minus four.
\[ 0 \]
Substitute into the denominator
Why: Two minus two.
\[ 0 \]
Identify the form
Why: Zero over zero.
Interpret
Why: Algebra is needed.
Figure (svg): The solution to Worked example substitution fails informatively shown as a ladder of expressions, one row per legal move
\[ \tfrac{0}{0}: \text{ indeterminate} \]
Verify: note what the form does not say
Why: It does not say the limit fails — the shared zero suggests a common factor, which cancelling will remove. The form identifies the situation rather than resolving it, which is exactly its usefulness.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1410-1411
Error analysis
A student evaluates a limit.
Annotate
On: \( \lim_{x\to 2}\frac{x^2-4}{x-2}=\frac{0}{0}=0 \)
The form is called indeterminate precisely because it does not determine the answer. Two expressions both giving zero over zero can have entirely different limits, which is why the algebra cannot be skipped.
Sorting
Check what it produces.
Sort into buckets
Sort each situation.
Faded example
Substituting two into a quotient.
Fill in the blanks
\frac00=\frac___}___}
Why: Both parts vanish at the point, giving the indeterminate form. That shared zero indicates a common factor, which factoring will expose and cancelling will remove.
Explain it
It is not just undefined.
Discussion prompt
Explain what indeterminate means here.
Hint: Compare two expressions with the same form.
Answer:
Two different expressions can both give zero over zero at a point and have different limits — so the form does not determine the answer.
That is what indeterminate means: the form is consistent with many outcomes, so more information is needed.
The information comes from the algebra — factoring, rationalising, simplifying — which reveals what the expression actually does near the point. A good explanation contrasts this with a nonzero number over zero, which does determine the outcome: no limit.
Section
Section 3
Concept
Zero over zero means the limit may exist and needs work. A nonzero number over zero means the outputs grow without bound and no limit exists.
The third and fourth rows are opposites and easy to confuse. Zero on top with a nonzero bottom is a perfectly ordinary limit of zero; zero on the bottom with a nonzero top is a failure.
| substitution gives | meaning |
|---|---|
| a number | that is the limit |
| zero over zero | indeterminate: do algebra |
| a nonzero number over zero | no limit; unbounded growth |
| zero over a nonzero number | the limit is zero |
Figure (svg): A contrast between an indeterminate form, which requires further work, and a genuinely undefined result, which means the limit fails
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1411-1413
Picture it
They look alike and mean opposite things.
Figure (svg): A contrast between an indeterminate form, which requires further work, and a genuinely undefined result, which means the limit fails
The caption is the summary worth carrying. One says keep working and the other says stop, and reading the wrong one wastes effort or abandons a limit that exists.
Worked example
Nonzero over zero.
\[ \text{Evaluate } \lim_{x\to 1}\frac{x+3}{x-1}. \]
Substitute into the numerator
Why: One plus three.
\[ 4 \]
Substitute into the denominator
Why: One minus one.
\[ 0 \]
Identify the form
Why: Nonzero over zero.
Conclude
Why: The outputs grow without bound.
Figure (svg): A contrast between an indeterminate form, which requires further work, and a genuinely undefined result, which means the limit fails
\[ \text{no limit} \]
Verify: check the behaviour near the point
Why: Just above 1 the denominator is small and positive, giving large positive outputs; just below it is small and negative, giving large negative ones. The outputs run away in both directions, confirming there is nothing to approach.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1412-1413
Sorting
The position of any zero decides.
Sort into buckets
Sort each result.
Worked example
An ordinary limit, not a failure.
\[ \text{Evaluate } \lim_{x\to 2}\frac{x-2}{x+3}. \]
Substitute into the numerator
Why: Two minus two.
\[ 0 \]
Substitute into the denominator
Why: Two plus three.
\[ 5 \]
Check the condition
Why: The bottom is nonzero.
Compute
Why: Zero over five.
\[ 0 \]
Figure (svg): The solution to Worked example zero on top shown as a ladder of expressions, one row per legal move
\[ 0 \]
Verify: contrast with the previous example
Why: Here the zero is on top and the bottom is nonzero, which is an ordinary division giving zero. The previous example had the zero on the bottom, which is a failure — the same digit in different positions means opposite things.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1413-1413
Trap
\[ \frac{0}{5}: \text{ a zero appears, so this is indeterminate} \]
React to the presence of a zero
Why: Its position is not checked.
Zero divided by a nonzero number is simply zero, and no work is needed.
Only zero over zero is indeterminate. A zero on top with a nonzero bottom is an ordinary division.
A zero on the bottom with a nonzero top is a failure, not an indeterminate form.
The position of the zero decides which of three situations applies, and all three are common.
Prediction
Substitution gives four over zero.
Predict first
What does that mean?
Correct: The limit does not exist.
Why: A nonzero numerator over a vanishing denominator means the outputs grow without bound near the point. No algebra will help, since there is no shared factor to cancel.
Faded example
Zero on top and five on the bottom.
Fill in the blanks
\frac0form=___, \text______
Why: The denominator being nonzero means the quotient property applies and the division is ordinary. Only a vanishing denominator causes any difficulty.
Explain it to yourself
Substitution can produce three quite different results.
Discussion prompt
Explain what each means.
Hint: Where is the zero?
Answer:
A number means that number is the limit, and nothing further is needed.
Zero over zero is indeterminate: the limit may exist and algebra is needed to find it, since the shared zero indicates a common factor.
A nonzero number over zero means the outputs grow without bound and no limit exists. A good explanation stresses that the last two look alike and mean opposite things, so the numerator has to be checked as well as the denominator.
Section
Section 4
Concept
Every technique for an indeterminate quotient does the same thing: expose the factor that vanishes in both parts and cancel it, after which substitution works.
Cancelling changes the function — the original had a hole and the simplified one does not — but it does not change the limit, because a limit ignores the point itself. That is what makes the technique legitimate.
Figure (svg): A card listing the three algebraic techniques for resolving an indeterminate quotient
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1413-1415
Picture it
Each suits a different shape of expression.
Figure (svg): A card listing the three algebraic techniques for resolving an indeterminate quotient
All three converge on the same goal. Reading the expression's shape decides which route is available, and the result is the same in every case.
Worked example
The commonest technique.
\[ \text{Evaluate } \lim_{x\to 2}\frac{x^2-4}{x-2}. \]
Substitute first
Why: To identify the form.
\[ \frac{0}{0} \]
Factor the numerator
Why: A difference of squares.
\[ (x - 2) (x + 2) \]
Cancel the shared factor
Why: Legitimate away from the point.
\[ x + 2 \]
Substitute again
Why: Into what remains.
\[ 4 \]
Figure (svg): A card listing the three algebraic techniques for resolving an indeterminate quotient
\[ 4 \]
Verify: check against the table from §12.1
Why: The table there converged on 4, which matches. The algebra confirms the numerical estimate exactly, which is the relationship between the two sections.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1414-1415
Matching
The expression's shape decides.
Match the pairs
Why: Each technique removes the same thing — the factor vanishing in both parts — by a route the expression's shape permits. Reading the shape is what selects the route.
Worked example
A root blocks factoring.
\[ \text{Evaluate } \lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}. \]
Substitute first
Why: Both parts vanish.
\[ \frac{0}{0} \]
Multiply by the conjugate
Why: Top and bottom.
Simplify the numerator
Why: A difference of squares.
Cancel and substitute
Why: The shared factor goes.
\[ \frac{1}{4} \]
Figure (svg): The solution to Worked example rationalise shown as a ladder of expressions, one row per legal move
\[ \tfrac{1}{4} \]
Verify: check the conjugate did its job
Why: Multiplying by the conjugate turned the numerator into x plus four minus four, which is x — cancelling with the denominator. The root was the obstacle and rationalising was what removed it.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1415-1415
Error analysis
A student simplifies and claims the functions are identical.
Annotate
On: \( \frac{x^2-4}{x-2}=x+2 \text{, so they are the same function} \)
The distinction matters conceptually even though it does not change the answer. Cancelling is valid for the limit precisely because a limit ignores the point where the two functions differ.
Prediction
The original and simplified expressions differ at one point.
Predict first
Why does that not affect the limit?
Correct: Because a limit ignores the point itself.
Why: The two expressions agree everywhere except at the point being approached, and the limit depends only on nearby behaviour. That is exactly why the cancellation is valid.
Faded example
After factoring a difference of squares.
Fill in the blanks
\frac22}}=x+___ \quad(x\ne 2)
Why: The shared factor cancels away from the point, leaving an expression that substitution handles. The restriction records that the two are not the same function at the point itself.
Explain it
It changes the function but not the limit.
Discussion prompt
Explain the distinction to a classmate.
Hint: Where do the two differ?
Answer:
The two expressions agree everywhere except at the point being approached, where the original is undefined and the simplified one is not.
But a limit depends only on nearby behaviour, deliberately ignoring the point itself. So the difference at that one point cannot affect it.
That is why cancelling is legitimate for a limit while the functions remain genuinely different. A good explanation notes that this is the §12.1 design decision doing real work, rather than being a technicality.
Section
Section 5
Concept
Once an indeterminate form appears, the expression's structure indicates which of the three techniques will expose the vanishing factor.
The last row matters: not every indeterminate form resolves with these techniques, and some limits genuinely do not exist. The techniques cover the cases that appear at this level, and calculus supplies more.
| what you see | what to try |
|---|---|
| polynomials top and bottom | factor and cancel |
| a square root in a difference | multiply by the conjugate |
| a fraction inside a fraction | combine and simplify |
| a piecewise definition | take one-sided limits |
| none of these | the limit may genuinely fail |
Figure (svg): A card listing the three algebraic techniques for resolving an indeterminate quotient
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1409-1415
Picture it
Read the shape to choose one.
Figure (svg): A card listing the three algebraic techniques for resolving an indeterminate quotient
All three remove the vanishing factor. Which one is available depends on whether the obstacle is a polynomial, a root, or nested fractions.
Worked example
The shape indicates the technique.
\[ \text{Which technique for } \lim_{x\to 0}\frac{\sqrt{x+9}-3}{x}? \]
Substitute
Why: To confirm the form.
\[ \frac{0}{0} \]
Look at the numerator
Why: A root minus a number.
Note factoring is blocked
Why: The root cannot factor.
Choose rationalising
Why: The conjugate clears it.
Figure (svg): A card listing the three algebraic techniques for resolving an indeterminate quotient
\[ \text{multiply by the conjugate} \]
Verify: carry it through
Why: The conjugate turns the numerator into x plus nine minus nine, which is x — cancelling with the denominator and leaving one over the sum of the root and 3, which is one sixth at the point. The technique worked because it was matched to the obstacle.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1413-1415
Sorting
The obstacle decides.
Sort into buckets
Sort each expression's obstacle.
Worked example
Different rules on each side.
\[ \text{A function is } x+1 \text{ below } 2 \text{ and } x^2 \text{ at or above. Find the limit at } 2. \]
Take the left-hand limit
Why: Using the lower rule.
\[ 3 \]
Take the right-hand limit
Why: Using the upper rule.
\[ 4 \]
Compare
Why: They differ.
Conclude
Why: No two-sided limit.
Figure (svg): The solution to Worked example a piecewise limit shown as a ladder of expressions, one row per legal move
\[ \text{no limit; }3\text{ and }4 \]
Verify: note why substitution alone was not enough
Why: Substituting requires knowing which rule applies, and the two sides use different rules — so the limit had to be taken one side at a time. A piecewise definition always calls for that, whatever the pieces look like.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1414-1415
Trap
\[ \text{try to factor a numerator containing a square root} \]
Apply the first technique regardless of shape
Why: The obstacle's nature is not considered.
A root does not factor, so the attempt goes nowhere.
Read the obstacle first. A root calls for rationalising and nested fractions for simplifying.
Each technique exposes the vanishing factor by a route the expression's shape permits.
A technique that does not fit will not fail loudly — it will simply produce no progress, which wastes time rather than signalling the problem.
Prediction
The rule changes at the point being approached.
Predict first
What do you do?
Correct: Take the two one-sided limits separately.
Why: Each side uses a different rule, so each one-sided limit is computed with its own. The two-sided limit then exists only if they agree, which is the §12.1 criterion applied here.
Faded example
Multiplying by the conjugate.
Fill in the blanks
(\sqrt9-3)(\sqrtx+3)=x+9-___=___
Why: The conjugate turns the difference into a difference of squares, which removes the root entirely. The result cancels with the denominator, which is why the technique resolves the form.
Explain it to yourself
Three techniques, one purpose.
Discussion prompt
Explain what they all achieve.
Hint: What is being removed?
Answer:
All three expose and remove the factor that vanishes in both the numerator and the denominator, which is what produced the indeterminate form.
Factoring does it directly, rationalising clears a root that was blocking the factoring, and simplifying a complex fraction untangles nested fractions to reveal the same thing.
Once the factor is gone, substitution works. A good explanation notes that the choice among them is about the obstacle's shape rather than about the goal, which is identical in every case.
Comparison
Fill the blanks from memory. All three are informative.
Comparison matrix
| a number | zero over zero | nonzero over zero | |
|---|---|---|---|
| meaning | that is the limit | indeterminate | no limit exists |
| what to do | nothing further | do algebra | stop |
| the limit | found | may still exist | does not exist |
| why | the function is well behaved | a shared vanishing factor | unbounded growth |
The middle and right columns look alike on the page and mean opposite things. Checking the numerator as well as the denominator is what separates them.
Pattern
Five steps, and the second decides everything after it.
Step 2 is where the section's real content lies. The three outcomes lead to three quite different next steps, and confusing the last two is the standard error.
Check
Indeterminate forms.
Check your understanding
Substitution gives zero over zero. What does that mean?
Answer: A
Why: The form determines nothing — two expressions producing it can have different limits. The shared zero indicates a common factor, which factoring or rationalising can remove.
Check
The other failure.
Check your understanding
Substitution gives five over zero. What does that mean?
Answer: A
Why: A nonzero numerator over a vanishing denominator means the outputs grow without bound near the point. There is no shared factor to cancel, so no algebra will help.
Check
Cancelling.
Check your understanding
Why is cancelling a shared factor valid when computing a limit?
Answer: A
Why: The original and simplified expressions differ only at the point being approached, and a limit depends only on nearby behaviour. That is exactly what makes the cancellation legitimate.
Real world
Every derivative computation begins with an indeterminate form.
Discussion prompt
A rate of change is a difference quotient. Why is it always zero over zero at the point of interest?
Hint: What happens to both parts as the interval shrinks?
Answer:
The numerator is a difference of outputs and the denominator a difference of inputs. As the interval shrinks, both go to zero.
So substituting directly always gives zero over zero — an indeterminate form, every time, by construction.
Which means the techniques of this section are not occasional tools but the standard first step of every derivative. Section 12.4 does exactly this, and calculus is largely the study of what that form resolves to in different cases.
Commit first
State your confidence along with your answer.
Predict first
Why is zero over zero called indeterminate rather than undefined?
Correct: Different expressions with that form can have different limits.
Why: The form is consistent with many outcomes, so it determines nothing on its own — which is what indeterminate means. A nonzero number over zero, by contrast, determines the outcome completely: no limit.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate wrote that a limit equals zero over zero. Explain the problem.
Hint: Is that a number?
Answer:
Zero over zero is not a number, so it cannot be an answer. It is a form that substitution produced.
And it is indeterminate: two different expressions can both give it and have entirely different limits, so it determines nothing.
What it does say is that the numerator and denominator share a vanishing factor, which factoring or rationalising can remove. A good explanation notes that this is a signal to keep working, and contrasts it with a nonzero number over zero, which is a signal to stop.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the two failures are confused, and they mean opposite things. The fourth is what turns a signal into an answer.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a decision tree starting from substitution, with its three outcomes and what each means. Beside it, list the four limit properties and mark the one carrying a condition. Underneath, work one factoring example and one rationalising example, noting in each why cancelling is legitimate.
If your decision tree separates zero over zero from a nonzero number over zero, and your cancelling note refers to the limit ignoring the point, the section's two ideas are on the page.
Recap
Five things, and the second is the decision that organises the rest.
| if you remember one thing | it should be this |
|---|---|
| about the properties | only the quotient one carries a condition |
| about substitution | try it first; all three outcomes are informative |
| about zero over zero | a signal to do algebra, not an answer |
| about cancelling | valid because a limit ignores the point itself |
Section 12.3 names the property that makes substitution work — continuity — and classifies the ways a function can fail to have it.
OpenStax, Precalculus, §12.2 Finding Limits: Properties of Limits §12.2, pp. 1404-1415 — everything on these slides traces back here
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