The basic limit laws and what they require, direct substitution for polynomials and rational functions, the five algebraic moves for resolving an indeterminate quotient, one-sided laws for piecewise functions, and the Squeeze Theorem with the fundamental trigonometric limits it delivers.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 2 — Limits
The Limit Laws
Objectives
Five outcomes. Together they turn limits from something you estimate into something you compute in one line.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 140-154 — the section these objectives are drawn from
Warm-up
Section 2.2 defined the limit and estimated several from tables. Estimating is slow and, as the oscillating example showed, it can be wrong.
Discussion prompt
You need the limit of x squared minus 3x as x approaches 2. Would you build a table, or is there something faster you already suspect works?
Hint: Try substituting and ask yourself why you believe the answer.
Answer:
\[ \lim_{x \to 2}(x^2 - 3x) = 4 - 6 = -2 \]
Substitution gives negative 2 immediately, and a table would agree. But why is substitution allowed? Nothing in the definition of a limit says the value at the point is relevant — Section 2.2 insisted at length that it is not.
The answer is that for polynomials the limit and the value happen to coincide, and this section proves it from a handful of basic laws rather than assuming it. Knowing why it works is what tells you the cases where it does not.
Concept
If two functions each have a limit at a point, then their sum, difference, product and quotient have the limits you would expect. From two trivial base cases — the limit of a constant and the limit of x itself — those laws build direct substitution for every polynomial.
the limit laws — If the limits of f and g at a point both exist, the limit of any sum, difference, constant multiple, product, power or root of them is obtained by applying that operation to the limits. The quotient law requires additionally that the denominator's limit not be zero.
\[ \lim_{x \to a}[f(x) + g(x)] = \lim_{x \to a}f(x) + \lim_{x \to a}g(x) \]
Read the hypothesis carefully: each law begins by assuming the pieces have limits. Every difficulty in this section comes from a case where that assumption fails — usually a denominator heading for zero.
Figure (svg): The limit laws listed, each showing that the limit passes through the operation
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 140-143
Section
Section 1
Concept
The limit of a constant is that constant, and the limit of x as x approaches a is a. Combining those two with the sum, product and constant multiple laws gives the limit of any polynomial by substitution, and the quotient law extends it to rational functions.
direct substitution — For a polynomial, the limit at a point equals the value at that point. For a rational function, the same holds at every point where the denominator does not vanish.
\[ \lim_{x \to a} p(x) = p(a); \qquad \lim_{x \to a}\frac{p(x)}{q(x)} = \frac{p(a)}{q(a)} \;\text{ if } q(a) \ne 0 \]
This is a theorem, not a convenience. It is exactly the statement that polynomials and rational functions are continuous wherever they are defined — which is what Section 2.4 will name and generalise.
Figure (svg): Direct substitution succeeding for a polynomial and failing for a quotient with a vanishing denominator
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 140-146 — the limit laws and direct substitution
Picture it
The same first move, two different outcomes.
Figure (svg): Direct substitution succeeding for a polynomial and failing for a quotient with a vanishing denominator
Substitution is always the first thing to try, because when it works it finishes the problem in one line. When it produces zero over zero it has not failed uselessly — it has told you which technique the expression needs.
Worked example
Example 2.14. Break it apart, then reassemble.
\[ \text{Evaluate } \lim_{x \to -3}(4x + 2). \]
Split with the sum law
Why: The limit of a sum is the sum of the limits.
\[ \lim 4 x + \lim 2 \]
Pull the constant out of the first
Why: The constant multiple law.
\[ 4 \lim x + \lim 2 \]
Apply the two base cases
Why: The limit of x is the point; the limit of a constant is itself.
\[ 4(-3) + 2 \]
Evaluate
Why: Negative 12 plus 2.
\[ -10 \]
Figure (svg): The solution to Worked example a limit by the laws shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to -3}(4x+2) = -10 \]
Verify: compare with direct substitution
Why: Substituting negative 3 into 4x plus 2 gives negative 10 immediately. The four-step derivation reaches the same place, which is exactly the point: the laws PROVE that substitution is legitimate here, so in practice one line suffices. Working through the laws once, as here, is what earns the right to skip them afterwards.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 143-144
Sorting
Check the denominator at the point.
Sort into buckets
Sort each limit.
The last one uses the root law, whose condition is that the radicand's limit be non-negative — here it is 9, so the law applies and the answer is 3. Both conditional laws were checked, and both passed.
Worked example
Checkpoint 2.14. Check the denominator first.
\[ \text{Evaluate } \lim_{x \to 2} \frac{2x^2 - 3x + 1}{x^3 + 4}. \]
Evaluate the denominator at the point
Why: Eight plus 4.
\[ 12,\text{ which is not } 0 \]
The quotient law therefore applies
Why: Its one condition is satisfied.
Substitute into the numerator
Why: Eight minus 6 plus 1.
\[ 3 \]
Form the quotient
Why: Three over 12.
\[ \frac{1}{4} \]
Figure (svg): The solution to Worked example a rational function where substitution works shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 2}\frac{2x^2-3x+1}{x^3+4} = \frac{3}{12} = \frac{1}{4} \]
Verify: confirm the condition was genuinely checked
Why: The denominator at 2 is 12, comfortably non-zero, so nothing indeterminate arose and the answer is exact. Checking the denominator BEFORE substituting is the habit worth forming: it takes one evaluation and it tells you immediately whether this is a one-line problem or a four-line one. Every difficulty in this section lives in the case where that check fails.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 145-146
Trap
\[ \lim_{x \to 2}\frac{x^2-4}{x-2} = \frac{\lim_{x \to 2}(x^2-4)}{\lim_{x \to 2}(x-2)} \]
Apply the quotient law without checking its condition
Why: The student splits the limit as usual.
\[ = \frac{0}{0} \quad \text{(the law never applied)} \]
The quotient law explicitly requires the denominator's limit to be non-zero. Here it is zero, so the law says nothing at all and the line is not a valid step.
\[ \frac{x^2-4}{x-2} = x+2 \;(x \ne 2) \;\Longrightarrow\; \lim_{x \to 2} = 4 \]
Check the hypothesis, and simplify first when it fails
Why: Cancel the common factor, which is legal at every input the limit uses.
Zero over zero is called indeterminate for a precise reason: expressions of that shape can have any limit whatever. The quotient of x squared minus 4 by x minus 2 tends to 4, while the quotient of x minus 2 by x squared minus 4 tends to one quarter, and the quotient of x minus 2 by the square of x minus 2 is unbounded. Same form, three different outcomes.
Fill the middle
Breaking a linear limit into base cases.
Fill in the blanks
\lim_-10(4x+2) = 4\lim____x + \lim____2 = 4(-3) + 2 = ___
Why: The sum and constant multiple laws reduce everything to the two base cases, and the answer is negative 10 — the same as direct substitution. This derivation is what licenses the shortcut for every polynomial.
Two truths and a lie
All three are about the laws.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Zero over zero determines nothing whatever: the quotient of x squared minus 4 by x minus 2 has limit 4, the reciprocal arrangement has limit one quarter, and other arrangements are unbounded or have no limit at all. That is exactly what indeterminate means, and it is why the form is a prompt for algebra rather than an answer.
Prediction
Commit before reasoning.
Predict first
Section 2.2 insisted a limit ignores the value at the point. So why may we substitute into a polynomial?
Correct: Because the laws prove the limit equals the value for polynomials.
\[ \lim_{x \to a}c = c, \quad \lim_{x \to a}x = a \;\Longrightarrow\; \lim_{x \to a}p(x) = p(a) \]
Why: The definition still ignores the point; what the laws establish is that for a polynomial the number the limit produces happens to coincide with the number substitution produces. That coincidence is a theorem built from the two base cases, and it is precisely the property Section 2.4 will call continuity. Recognising it as a proved coincidence rather than a relaxed rule is what tells you where it stops — at any point where a denominator vanishes, and at every jump or corner in Chapter 3.
Section
Section 2
Concept
When substitution gives zero over zero in a rational expression, the numerator and denominator share a factor that vanishes at the point. Cancelling it produces a new expression agreeing with the old at every input the limit uses, and substitution then works.
indeterminate form — An expression such as zero over zero whose limit is not determined by the form alone. Different functions producing the same form can have entirely different limits, so the form is an instruction to simplify rather than an answer.
\[ \frac{x^2-4}{x-2} = x + 2 \quad \text{for } x \ne 2 \]
The cancellation is legitimate for exactly the reason Section 2.2 laboured: the two expressions differ only at the single point being approached, and the limit never uses that point.
Figure (svg): The five algebraic moves for resolving an indeterminate quotient, each with the shape that triggers it
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 146-149 — additional limit evaluation techniques
Picture it
What to do when substitution gives zero over zero.
Figure (svg): The five algebraic moves for resolving an indeterminate quotient, each with the shape that triggers it
The shape of the expression selects the move. A quotient of polynomials wants factoring; a square root wants the conjugate; a fraction inside a fraction wants combining; a binomial power wants expanding.
Worked example
Example 2.17. The commonest case by far.
\[ \text{Evaluate } \lim_{x \to 3} \frac{x^2 - 3x}{2x^2 - 5x - 3}. \]
Substitute to identify the form
Why: Numerator 0, denominator 18 minus 15 minus 3, also 0.
\[ \frac{0}{0} \]
Factor the numerator
Why: Take out the common x.
\[ x(x - 3) \]
Factor the denominator
Why: It has x minus 3 as a factor.
\[ (x - 3) (2 x + 1) \]
Cancel the shared factor
Why: Legal for every x other than 3.
\[ \frac{x}{2 x + 1} \]
Substitute into the simplified form
Why: Three over 7.
\[ \frac{3}{7} \]
Figure (svg): The solution to Worked example factor and cancel shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 3}\frac{x^2-3x}{2x^2-5x-3} = \frac{3}{7} \]
Verify: check the factorisation by expanding
Why: Multiplying x minus 3 by 2x plus 1 gives 2x squared plus x minus 6x minus 3, which is 2x squared minus 5x minus 3 — the original denominator. A numerical check confirms the limit: at x equal to 3.001 the original expression evaluates to about 0.42863, and 3 over 7 is 0.428571. That the shared factor was x minus 3 is not luck: whenever substitution gives 0/0 in a rational function, x minus a is guaranteed to divide both parts, by the Factor Theorem.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 147-148
Matching
The shape tells you the technique.
Match the pairs
Why: Every one of these begins as zero over zero and ends with a common factor cancelled — the four moves differ only in how they manufacture that factor. Seeing them as one technique with four preparations is more useful than memorising four recipes.
Worked example
The shape every derivative in Chapter 3 produces.
\[ \text{Evaluate } \lim_{h \to 0} \frac{(2+h)^2 - 4}{h}. \]
Substitute to identify the form
Why: Four minus 4, over 0.
\[ \frac{0}{0} \]
Expand the square
Why: Four plus 4h plus h squared.
\[ \frac{4 + 4 h + h ^{2} - 4}{h} \]
Simplify the numerator
Why: The constants cancel.
\[ \frac{4 h + h ^{2}}{h} \]
Factor out h and cancel
Why: Legal because h is not 0.
\[ 4 + h \]
Substitute
Why: As h approaches 0.
\[ 4 \]
Figure (svg): The solution to Worked example expanding a binomial power shown as a ladder of expressions, one row per legal move
\[ \lim_{h \to 0}\frac{(2+h)^2-4}{h} = 4 \]
Verify: recognise what has just been computed
Why: This is exactly the derivative of x squared at x equal to 2, and the answer 4 matches the rule that the derivative of x squared is 2x. Section 2.1's secant slopes were the same computation with a different letter. The cancellation of the constants in step three is the crucial move, and it is why the numerator always has h as a factor — the constant terms are designed to cancel, which is what makes the difference quotient tractable at all.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 148-149
Error analysis
A student simplifies an indeterminate quotient.
Annotate
On: \( \frac{x^2 - 3x}{2x^2 - 5x - 3} = \frac{x^2 - 3x}{2x^2 - 5x - 3} \;\to\; \frac{-3x}{-5x-3} \quad \text{(cancelling the squares)} \)
Factor completely first, then cancel only whole factors. The one-input numerical test costs ten seconds and catches every version of this error, which is the most common algebraic mistake in the entire chapter.
Fill the middle
The rational limit from the worked example, with both parts factored.
Fill in the blanks
\frac7___ = \frac______ \;\Longrightarrow\; \lim____ = \frac______}
Why: Substituting 3 into the simplified form gives 3 over 7. The cancellation was legal because the two expressions agree at every input except 3, and the limit uses no others.
Ranking
Evaluating a limit that turns out indeterminate.
Put in order
Why: Step a is never wasted: it either finishes the problem or diagnoses it. Step d is where the legitimacy lies, and it is worth stating explicitly the first several times — the cancellation is valid precisely because the limit never evaluates at the point.
Prediction
Commit before reasoning.
Predict first
For a rational function giving 0/0 at x = a, why must x - a divide both parts?
Correct: By the Factor Theorem — a polynomial vanishing at a has x minus a as a factor.
\[ p(a) = 0 \;\Longrightarrow\; p(x) = (x-a)q(x) \]
Why: Getting 0/0 means both polynomials vanish at a, and the Factor Theorem from Section 1.2 then guarantees each has x minus a as a factor. So the cancellation is always available for rational functions; it is never a matter of luck. This also predicts what happens when the factor appears more than once in the denominator — after cancelling once, an unbounded expression may remain, and the limit is infinite rather than finite.
Section
Section 3
Concept
When a square root or a nested fraction blocks the cancellation, an algebraic preparation exposes it. Multiplying by a conjugate turns a difference of roots into a difference of squares; combining inner fractions turns a compound fraction into an ordinary one.
conjugate — For an expression of the form a minus the square root of b, the conjugate is a plus that root. Multiplying the two gives a difference of squares, which removes the root.
\[ (\sqrt{x}-2)(\sqrt{x}+2) = x - 4 \]
Both techniques are the same idea as factoring: the goal is always a common factor that cancels. What changes is the algebra needed to make it visible.
Figure (svg): Rationalising a numerator to remove an indeterminate form
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 149-152 — rationalising and complex fractions
Picture it
One indeterminate quotient, five lines.
Figure (svg): Rationalising a numerator to remove an indeterminate form
Multiplying by the conjugate over itself changes nothing, since that fraction is 1. What it does is move the root from a place where it blocks cancellation to a place where it is harmless.
Worked example
Example 2.18. The conjugate does all the work.
\[ \text{Evaluate } \lim_{x \to 4} \frac{\sqrt{x}-2}{x-4}. \]
Substitute to identify the form
Why: Two minus 2 over 0.
\[ \frac{0}{0} \]
Multiply top and bottom by the conjugate
Why: This is multiplying by 1, so nothing changes.
\[ \times(\sqrt{x} + 2) / (\sqrt{x} + 2) \]
Simplify the numerator as a difference of squares
Why: The roots disappear.
\[ \frac{x - 4}{(x - 4) (\sqrt{x} + 2)} \]
Cancel
Why: Legal because x is not 4.
\[ \frac{1}{\sqrt{x} + 2} \]
Substitute
Why: Two plus 2 in the denominator.
\[ \frac{1}{4} \]
Figure (svg): The solution to Worked example rationalising a numerator shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 4}\frac{\sqrt{x}-2}{x-4} = \frac{1}{4} \]
Verify: check numerically and note what it computes
Why: At x equal to 4.01 the original gives about 0.249844, and one quarter is 0.25. This limit is the derivative of the square root function at 4, and the rule from Chapter 3 gives one over twice the root of 4, which is one quarter — agreeing exactly. Note the conjugate was applied to the NUMERATOR because that is where the root sat; applying it to the denominator instead would have achieved nothing.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 150-151
Fill the middle
Rationalising the numerator from the worked example.
Fill in the blanks
(\sqrtx - 4-2)(\sqrt___+2) = ___
Why: The product is x minus 4, which is exactly the denominator — so it cancels. That the conjugate manufactures precisely the factor needed is not luck; it is why the technique works on this shape every time.
Worked example
Checkpoint 2.18. Tidy the inside before touching the outside.
\[ \text{Evaluate } \lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}. \]
Substitute to identify the form
Why: One half minus one half over 0.
\[ \frac{0}{0} \]
Combine the two inner fractions
Why: Common denominator 2x.
\[ \frac{2 - x}{2 x} \]
Rewrite the whole compound fraction
Why: Divide by x minus 2, that is multiply by its reciprocal.
\[ \frac{2 - x}{2 x(x - 2)} \]
Recognise the sign relationship
Why: Two minus x is the negative of x minus 2.
\[ -\frac{x - 2}{2 x(x - 2)} \]
Cancel and substitute
Why: Leaving negative one over 2x.
\[ -\frac{1}{4} \]
Figure (svg): The solution to Worked example a compound fraction shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 2}\frac{\frac1x - \frac12}{x-2} = -\frac{1}{4} \]
Verify: confirm the sign, which is where this one goes wrong
Why: At x equal to 2.01 the original gives about negative 0.2488, so the answer is genuinely negative. The sign comes from step four: 2 minus x is the negative of x minus 2, and missing that flip gives positive one quarter. This is the derivative of one over x at 2, and the rule gives negative one over x squared, which is negative one quarter — the negative sign is expected, because the reciprocal function is decreasing.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 151-152
Trap
\[ \lim_{x \to 4}\frac{\sqrt{x}-2}{x-4} \]
Multiply top and bottom by the conjugate of the DENOMINATOR
Why: The student writes x plus 4 and multiplies through.
\[ \frac{(\sqrt{x}-2)(x+4)}{(x-4)(x+4)} = \frac{(\sqrt{x}-2)(x+4)}{x^2-16} \]
The root is still in the numerator and both parts still vanish at 4. Nothing has been achieved and the expression is now more complicated.
\[ \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)} = \frac{x-4}{(x-4)(\sqrt{x}+2)} \]
Rationalise the part that CONTAINS the root
Why: The conjugate is a tool for removing a root, so it belongs where the root is.
Here the root sits in the numerator, so the numerator is what gets rationalised — which is the opposite of the habit formed in algebra, where denominators are rationalised for presentation. In limits the goal is cancellation, not tidiness, and that decides which part to treat.
Sorting
Look at what is blocking the cancellation.
Sort into buckets
Sort each indeterminate quotient.
Every route ends at the same place: a common factor cancelled, then substitution. Recognising the shape is the only decision to make, and it takes a glance rather than a calculation.
Two truths and a lie
All three are about these techniques.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it comes from a habit formed elsewhere. In algebra you rationalise denominators for presentation; in limits you rationalise wherever the root sits, because the aim is to create a cancellable factor rather than to tidy the expression. In the worked example the root was in the numerator, and treating the denominator would have accomplished nothing.
Prediction
Commit before reasoning.
Predict first
What does multiplying by the conjugate actually accomplish?
Correct: It converts a difference of roots into a difference of squares, manufacturing the cancellable factor.
\[ (\sqrt{a}-b)(\sqrt{a}+b) = a - b^2 \quad \text{- the root is gone} \]
Why: The form is still zero over zero immediately after multiplying — nothing about the value has changed, since the conjugate fraction equals 1. What has changed is that the numerator is now a polynomial sharing a visible factor with the denominator, and cancelling that factor is what removes the indeterminacy. The limit itself is never altered by any of these moves, which is precisely why they are allowed.
Section
Section 4
Concept
Every limit law holds equally for one-sided limits. So a piecewise function is handled by applying the laws to each rule on its own side, then comparing the two results with the theorem from Section 2.2.
one-sided limit laws — Each of the limit laws holds with the two-sided limits replaced throughout by left-hand limits, or throughout by right-hand limits. This is what makes piecewise functions computable rather than merely describable.
\[ \lim_{x \to a^-}[f+g] = \lim_{x \to a^-}f + \lim_{x \to a^-}g \]
Absolute values are piecewise functions in disguise, and are handled the same way: split at the input where the inside expression changes sign, and treat each side with its own formula.
Figure (svg): The five algebraic moves for resolving an indeterminate quotient, each with the shape that triggers it
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 152-154 — limits of piecewise functions
Picture it
Nothing new is needed — only applied twice.
Figure (svg): The five algebraic moves for resolving an indeterminate quotient, each with the shape that triggers it
A one-sided limit is evaluated exactly like a two-sided one, using the rule that governs that side. The only extra step is comparing the two answers at the end.
Worked example
Example 2.24. Two computations, then a comparison.
\[ \text{For } f(x) = \begin{cases} 4x - 3 & x < 2 \\ x^2 & x \ge 2 \end{cases} \text{ find } \lim_{x \to 2} f(x). \]
Compute the left-hand limit with the first rule
Why: Inputs below 2 use 4x minus 3.
\[ 4(2) - 3 = 5 \]
Compute the right-hand limit with the second rule
Why: Inputs above 2 use the square.
\[ 2 ^{2} = 4 \]
Compare the two
Why: Five against 4.
Apply the theorem
Why: Both exist but disagree.
Figure (svg): The solution to Worked example a piecewise limit shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 2^-}f = 5 \ne 4 = \lim_{x \to 2^+}f \]
Verify: check the value at 2 separately
Why: The second rule carries the equals sign, so f of 2 is 4 — which matches the right-hand limit only. This is a jump discontinuity of size 1, and all three quantities differ in the informative way: the left limit is 5, the right limit and the value are both 4. Section 2.4 will use exactly this pattern to classify the break.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 152-153
Fill the middle
The left-hand limit of the piecewise function from the worked example.
Fill in the blanks
\lim_5f(x) = \lim____(4x-3) = ___
Why: Inputs below 2 use the linear rule, giving 5. The right-hand limit uses the squaring rule and gives 4, so the two-sided limit does not exist.
Worked example
Checkpoint 2.24. Split where the inside changes sign.
\[ \text{Evaluate } \lim_{x \to 3} \frac{|x-3|}{x-3}. \]
Find where the inside expression changes sign
Why: At the point being approached.
\[ x - 3\text{ changes sign at } 3 \]
Rewrite for inputs above 3
Why: The inside is positive, so the absolute value is itself.
\[ \frac{x - 3}{x - 3} = 1 \]
Rewrite for inputs below 3
Why: The inside is negative, so the absolute value flips it.
\[ -\frac{x - 3}{x - 3} = -1 \]
Compare the two one-sided limits
Why: One and negative 1.
Figure (svg): The solution to Worked example a limit with an absolute value shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 3^-} = -1, \quad \lim_{x \to 3^+} = 1 \]
Verify: confirm the function is constant on each side
Why: For every input other than 3 the expression is exactly 1 or exactly negative 1 — there is no approach at all, just two constant values. So the one-sided limits are immediate, and the jump has size 2. This is the standard example of a limit failing by a jump, and every absolute value over its own inside behaves the same way at the sign change.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 153-154
Error analysis
A student evaluates a piecewise limit.
Annotate
On: \( f(x) = \begin{cases}4x-3 & x<2\\ x^2 & x \ge 2\end{cases} \;\Longrightarrow\; \lim_{x \to 2}f(x) = 2^2 = 4 \)
The rule that owns the point is irrelevant to the limit — it settles the VALUE. A piecewise limit always requires two computations, and reporting one of them is answering half the question.
Matching
Split where the inside changes sign.
Match the pairs
Why: An absolute value is a piecewise function with the split at the input where its inside vanishes. Rewriting it that way before taking any limit turns an awkward expression into two ordinary ones, and it is the only reliable way to handle these.
Sorting
Compute both sides, then compare.
Sort into buckets
Sort each function at the point named.
The absolute value of x at 0 is the instructive one: both sides approach 0, so the limit exists and equals 0 even though the graph has a sharp corner. A corner breaks the DERIVATIVE, not the limit — a distinction Section 3.2 will make much of.
Prediction
Commit before reasoning.
Predict first
Does the sum law hold for left-hand limits?
Correct: Yes. Every law holds verbatim with all limits replaced by left-hand ones, or all by right-hand ones.
\[ \lim_{x \to a^-}[f g] = \left(\lim_{x \to a^-}f\right)\!\left(\lim_{x \to a^-}g\right) \]
Why: The laws are statements about approaching, and restricting to one side changes nothing about the argument. This matters practically: it is what allows a piecewise function's one-sided limit to be computed by ordinary substitution into the relevant rule, rather than by returning to the definition. If the laws did not survive the restriction, every piecewise limit would need a table.
Section
Section 5
Concept
If a function is caught between two others near a point, and those two have the same limit there, then the trapped function must have that limit too. It has nowhere else to go.
Squeeze Theorem — If g of x is at most f of x is at most h of x for all inputs near a except possibly a itself, and g and h both have limit L at a, then f has limit L at a as well.
\[ g(x) \le f(x) \le h(x) \;\text{ and }\; g, h \to L \;\Longrightarrow\; f \to L \]
This is the tool for functions the algebraic moves cannot touch, especially those that oscillate. The oscillation is never resolved; it is simply confined to a gap that closes to nothing.
Figure (svg): A function trapped between two others that meet at a point, forcing its limit
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 154-154 — the Squeeze Theorem
Picture it
A wildly oscillating function inside a closing envelope.
Figure (svg): A function trapped between two others that meet at a point, forcing its limit
The middle curve crosses the horizontal axis infinitely often near the origin and has no pattern at all. The envelope closes to a point, and that is enough to force the limit to 0 without ever describing the oscillation.
Worked example
Example 2.25. Bound the oscillating part, then multiply.
\[ \text{Evaluate } \lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right). \]
Bound the oscillating factor
Why: A sine never leaves the interval from negative 1 to 1.
\[ -1 \le \sin(\frac{1}{x}) \le 1 \]
Multiply through by the non-negative factor
Why: Multiplying by x squared preserves the inequalities.
\[ -x ^{2} \le x ^{2} \sin(\frac{1}{x}) \le x ^{2} \]
Take limits of the two bounds
Why: Both are polynomials.
\[ \text{both tend to } 0 \]
Apply the theorem
Why: The trapped function is forced to the same limit.
\[ \lim = 0 \]
Figure (svg): The solution to Worked example squeezing an oscillation shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0} x^2\sin\!\left(\tfrac1x\right) = 0 \]
Verify: check that no algebraic move would have worked
Why: The expression is not a quotient, so there is nothing to factor, rationalise or combine; and substitution gives 0 times an undefined oscillation, which is meaningless. The Squeeze Theorem is the only available tool. Note also that multiplying the inequality by x squared was legal precisely because x squared is never negative — multiplying an inequality by a negative quantity would have reversed it, and that is the one place this argument can go wrong.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 154-154
Fill the middle
Bounding the oscillating product from the worked example.
Fill in the blanks
-x^2 \le x^2\sin\!\left(\tfrac1x\right) \le x^2, \quad \text0 ___
Why: Both bounds tend to 0, so the envelope closes to a single point and the trapped function is forced to 0. Had the bounds been negative 1 and 1, nothing would follow.
Worked example
The most important limit in the course, from pure geometry.
\[ \text{Show that } \lim_{t \to 0} \frac{\sin t}{t} = 1. \]
Compare three areas on the unit circle
Why: An inscribed triangle, the sector, and a circumscribed triangle.
\[ \sin t / 2 < \frac{t}{2} < \tan t / 2 \]
Multiply through by 2 and divide by sine, positive for small positive t
Why: The inequalities survive.
\[ 1 < t / \sin t < 1 / \cos t \]
Invert, which reverses the inequalities
Why: Taking reciprocals of positives.
\[ \cos t < \frac{\sin t}{t} < 1 \]
Take limits of the two bounds
Why: Cosine approaches 1, and 1 is constant.
\[ \text{both tend to } 1 \]
Apply the theorem
Why: The quotient is squeezed.
\[ \lim = 1 \]
Figure (svg): The unit circle argument bounding sine of t between t and the tangent, giving the fundamental limit
\[ \lim_{t \to 0}\frac{\sin t}{t} = 1 \]
Verify: confirm the role of radians and check the even symmetry
Why: The middle step used the sector's area as t over 2, which is true only when t is in radians — in degrees the sector area carries a factor of pi over 180 and the limit becomes that instead of 1. The argument was made for small positive t, and the quotient is an even function since sine is odd, so the negative side follows immediately. This limit is what Section 3.5 differentiates sine with, and it is the reason the whole course works in radians.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 154-154
Trap
\[ -1 \le \sin\!\left(\tfrac1x\right) \le 1 \]
Apply the Squeeze Theorem to the sine alone
Why: The student bounds it and concludes something about its limit at 0.
\[ \text{so } \lim_{x \to 0}\sin\!\left(\tfrac1x\right) \text{ exists} \quad \text{(wrong)} \]
The two bounds are negative 1 and 1, which do not meet. The theorem requires them to share a limit, and these do not.
\[ -x^2 \le x^2\sin\!\left(\tfrac1x\right) \le x^2, \;\text{ both bounds } \to 0 \]
Check that the two bounds have the SAME limit
Why: Boundedness alone proves nothing; the envelope must close.
The sine of one over x is bounded and has no limit at all at 0 — Section 2.2 established that. Multiplying by x squared is what makes the envelope close, and it is the factor doing all the work. Being trapped is not enough; being trapped in a gap that shrinks to a point is the hypothesis.
Two truths and a lie
All three are about the Squeeze Theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the misconception the theorem is most often misapplied through. The sine of one over x is bounded between negative 1 and 1 and has no limit whatever at 0. Boundedness constrains how large a function gets; it says nothing about whether it settles. The theorem needs a shrinking envelope, not merely a bounded one.
Ranking
Establishing a limit by squeezing.
Put in order
Why: Step c's sign check is the one that bites: multiplying an inequality by a negative quantity reverses it, so the bounds would swap. Step d is the hypothesis that is most often assumed rather than verified, and it is where the misapplication in the trap slide went wrong.
Prediction
Commit before reasoning.
Predict first
In the area argument, where do radians become essential?
Correct: In the sector's area, which is t over 2 only in radians.
\[ \text{sector area} = \tfrac{1}{2}r^2\theta \quad \text{(radians only)} \]
Why: A sector of a unit circle with central angle t has area t over 2 precisely because radian measure is defined by arc length — that is the arc-length relation from Section 1.3 doing its work. In degrees the sector area would be pi t over 360, and the squeeze would deliver pi over 180 rather than 1. The triangle areas and the reciprocals are unit-independent; the sector is not. This single step is the entire reason calculus insists on radians.
Comparison
Fill the blanks. The shape of the expression selects the move.
Comparison matrix
| The shape | The move | What it produces |
|---|---|---|
| Quotient of polynomials, 0/0 | factor both parts | a common factor x - a to cancel |
| A square root in a difference | multiply by the conjugate | a difference of squares, root removed |
| A fraction inside a fraction | combine over a common denominator | an ordinary quotient |
| A binomial power over h | expand, then factor out h | the constant terms cancel, leaving h as a factor |
| A bounded oscillation | squeeze between two bounds | a forced limit, oscillation unresolved |
The first four all end in the same place — a cancelled common factor — and differ only in the preparation. The fifth is genuinely different, and it is the one to reach for when nothing cancels.
Pattern
Given any limit to evaluate.
For a piecewise function or an absolute value, split first and then run this whole procedure separately on each side, comparing the two answers at the end.
Stewart, Calculus: Early Transcendentals 8e, §2.3 Calculating Limits Using the Limit Laws §2.3, pp. 95-103
Check
Substitution, when the denominator allows it.
Check your understanding
Evaluate the limit of (2x^2 - 3x + 1)/(x^3 + 4) as x approaches 2.
Answer: A
Why: The denominator at 2 is 12, which is non-zero, so substitution gives 3 over 12.
Check
An indeterminate quotient. Factor first.
Check your understanding
Evaluate the limit of (x^2 - 3x)/(2x^2 - 5x - 3) as x approaches 3.
Answer: A
Why: Both parts factor with x - 3; cancelling leaves x/(2x+1), which gives 3/7 at x = 3.
Check
Squeezing. Check the envelope closes.
Check your understanding
Evaluate the limit of x^2 sin(1/x) as x approaches 0.
Answer: A
Why: The product is trapped between -x^2 and x^2, both of which tend to 0.
Real world
A signal processing engineer models a damped oscillation as the product of a decaying amplitude and an oscillating carrier, and needs to know the signal's behaviour as time approaches a switch-on instant.
Discussion prompt
The signal near the instant behaves like t squared times the sine of one over t. Explain why the engineer can guarantee the signal settles to zero without ever describing its oscillation, and what would change if the amplitude decayed only like a constant.
Hint: The whole argument is the Squeeze Theorem, applied to a physical quantity.
Answer:
The carrier oscillates without limit — its frequency grows without bound as t approaches the instant, and no description of its behaviour is available or needed.
\[ -t^2 \le t^2\sin\!\left(\tfrac1t\right) \le t^2 \]
The amplitude envelope is what settles the question. Both bounds tend to 0, so the signal is trapped in a gap that closes, and it must go to 0. The engineer gets a guarantee about the signal without any information about the oscillation, which is exactly what makes the Squeeze Theorem valuable in practice.
If the amplitude were constant, the bounds would be negative 1 and 1 — an envelope that never closes. The theorem would say nothing, and correctly so: the signal would keep oscillating at full strength, with no settling value. The design consequence is real, and it is why damping is engineered in deliberately: it is the decaying envelope, not the oscillation, that guarantees the signal converges.
\[ \text{envelope closes} \Rightarrow \text{convergence guaranteed}; \quad \text{envelope constant} \Rightarrow \text{no conclusion} \]
Commit first
Answer, then rate your confidence honestly.
Predict first
A limit evaluates to 0/0. What does that tell you?
Correct: Nothing yet. The form is indeterminate, and the expression must be simplified first.
\[ \tfrac{0}{0} \text{ can be } 4, \; \tfrac14, \; \infty, \text{ or nonexistent} \]
Why: Expressions of the form zero over zero can have any limit whatever, or none: the quotient of x squared minus 4 by x minus 2 tends to 4, the reciprocal arrangement tends to one quarter, and the quotient of x minus 2 by its own square is unbounded. Since the form alone is compatible with all of these, it determines nothing. What it does reliably tell you is that a common factor is present and can be cancelled — which is an instruction, not an answer.
Explain it
They object that cancelling the factor changes the function, so the answer must be wrong.
Discussion prompt
In four sentences or fewer, explain why cancelling is legitimate even though it changes the function's domain.
Hint: Ask which inputs the limit actually uses.
Answer:
They are right that it changes the function: the quotient of x squared minus 4 by x minus 2 is undefined at 2, while x plus 2 is perfectly fine there. The two functions genuinely differ, at exactly one input.
But that input is the only one the limit never uses — the definition says near a, not equal to a. So the two functions agree everywhere the limit looks, and therefore have the same limit. The cancellation is not an approximation; it is an exact replacement on the domain that matters.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For choosing the move, look for what is BLOCKING the cancellation — a root wants a conjugate, a nested fraction wants combining. For rationalising, remember the conjugate goes where the root is, and check the sign at one nearby input. For piecewise, write the condition beside each one-sided limit before evaluating. For squeezing, verify the two bounds share a limit before doing anything else. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top of the page list the seven limit laws, marking the two that carry conditions and writing those conditions beside them. Under that, write the two base cases and show in three lines how they give direct substitution for any polynomial. In the middle, make a four-column table of the algebraic moves: for each, write the shape that triggers it, one example, and the common factor it produces. Work each of your four examples through completely. Below, take the piecewise function that is 4x minus 3 below 2 and x squared from 2 up, compute both one-sided limits, state whether the two-sided limit exists, and separately state the value at 2. At the bottom, draw the squeeze picture for x squared times the sine of one over x, marking the envelope, and beside it write the three-line area argument on the unit circle that gives the limit of sine t over t. In a margin, write one sentence saying why zero over zero is an instruction rather than an answer.
If your area argument does not use the fact that a unit-circle sector has area t over 2, look again — that step is where radians enter, and it is the only reason the answer is 1 rather than pi over 180.
Recap
Five things, and between them they evaluate almost every limit in the rest of the course.
| If you see | Then |
|---|---|
| A non-vanishing denominator | Substitute; you are done |
| 0/0 with polynomials | Factor: x - a divides both |
| 0/0 with a root | Multiply by the conjugate |
| A fraction inside a fraction | Combine over a common denominator |
| A binomial power over h | Expand; the constants cancel |
| A bounded oscillation times something small | Squeeze |
| An absolute value | Split at the sign change and do both sides |
Section 2.4 names the property that made substitution work. Continuity is precisely the condition that the limit and the value agree — and the theorems it unlocks, especially the Intermediate Value Theorem, are what let you assert a solution exists before you can find it.
OpenStax Calculus Volume 1, §2.3 The Limit Laws §2.3, pp. 140-154 — everything on these slides traces back here
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