The intuitive definition of a limit and the crucial fact that it ignores the value at the point, estimating limits from tables and graphs and the ways a table can mislead, the three ways a limit fails to exist, one-sided limits and the theorem connecting them to the two-sided limit, and infinite limits with vertical asymptotes.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 2 — Limits
The Limit of a Function
Objectives
Five outcomes. The first is a single sentence, and getting it exactly right makes the other four straightforward.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 116-139 — the section these objectives are drawn from
Warm-up
Section 2.1 computed secant slopes at inputs approaching a point but never equal to it, and read off where they were heading. That manoeuvre now gets a name and a definition.
Discussion prompt
The expression x squared minus 1 over x minus 1 is undefined at x equal to 1. Does that stop us from saying where it is heading as x approaches 1?
Hint: The table in Section 2.1 never used x equal to 1 and still gave a clear answer.
Answer:
\[ \frac{x^2-1}{x-1} = x + 1 \quad \text{for every } x \ne 1 \]
Not in the least. Every input in the table was different from 1, and every value computed was legitimate. The values closed on 2 without the point itself ever being consulted.
This is the central feature of the limit, not a technicality to work around: the limit is determined entirely by inputs near the point, and never by the point itself. Once that is genuinely accepted, the rest of the section follows.
Concept
The limit of a function as the input approaches a value is the single number the outputs get arbitrarily close to, provided the inputs are close enough to that value but not equal to it. The function need not be defined at the point, and if it is, its value there has no bearing on the limit.
limit of a function — If the values of a function get arbitrarily close to a single number L whenever the input is sufficiently close to a, but not equal to a, then L is the limit of the function as the input approaches a.
\[ \lim_{x \to a} f(x) = L \]
The three graphs above differ only at the single input 2 — one is continuous there, one has a hole, one has its value placed somewhere else entirely. All three have the same limit, because the limit never inspects that input.
Figure (svg): Three functions with the same limit at a point: one continuous, one with a hole, one with a misplaced value
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 116-119
Section
Section 1
Concept
To say the limit is L means the outputs can be forced as close to L as anyone demands, by taking the input close enough to a. The phrase but not equal to a is not decoration; it is what allows a limit to exist where the function does not.
approaching a value — The outputs approach L if, for any tolerance however small, all inputs sufficiently near a but different from a produce outputs within that tolerance of L.
\[ \lim_{x \to 2} (x+2) = 4 \quad \text{whether or not } f(2) \text{ is defined} \]
The whole of Section 2.1's difficulty dissolves here. The secant slope is undefined at the point of tangency and has a perfectly good limit there, and the limit is what we call the tangent slope.
Figure (svg): Three functions with the same limit at a point: one continuous, one with a hole, one with a misplaced value
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 116-121 — the intuitive definition of a limit
Picture it
Identical everywhere except at the single input 2.
Figure (svg): Three functions with the same limit at a point: one continuous, one with a hole, one with a misplaced value
The middle graph has a hole and the right one has its value misplaced. Neither changes what the curve is heading toward, and the limit is 4 in all three cases.
Worked example
Example 2.5. The function is undefined exactly where we are looking.
\[ \text{Evaluate } \lim_{x \to 1} \frac{x^2 - 1}{x - 1}. \]
Try direct substitution and note what happens
Why: Both numerator and denominator vanish.
\[ \frac{0}{0},\text{ indeterminate} \]
Factor the numerator
Why: A difference of squares.
\[ (x - 1) (x + 1) / (x - 1) \]
Cancel, which is legal because x is not 1
Why: The limit never uses the point itself.
\[ = x + 1\text{ for } x \ne 1 \]
Evaluate the simplified expression at the point
Why: The simplified form is defined there.
\[ \lim = 2 \]
Figure (svg): A two-sided table of values closing in on four from both directions
\[ \lim_{x \to 1} \frac{x^2-1}{x-1} = 2 \]
Verify: check with a two-sided table
Why: At 0.999 the value is 1.999 and at 1.001 it is 2.001, so both sides close on 2. Note carefully that the cancellation did not change the function on the domain that matters: the original and x plus 1 agree at every input except 1, and the limit only looks at inputs other than 1. The two functions have different domains and identical limits, which is exactly what makes the technique legitimate rather than a sleight of hand.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 119-120
Sorting
One asks about nearby inputs, the other about the point itself.
Sort into buckets
For the function equal to x + 2 except at 2, where it equals 1, sort each statement.
The last item is the sharpest test of understanding. That the limit survives any redefinition at the point is not a quirk — it is the property that lets a limit exist where a function does not, which is the entire reason Chapter 2 comes before Chapter 3.
Worked example
Checkpoint 2.5. The function is defined, and defined wrongly.
\[ \text{For } g(x) = \begin{cases} x + 2 & x \ne 2 \\ 1 & x = 2 \end{cases} \text{ find } \lim_{x \to 2} g(x) \text{ and } g(2). \]
Read the value at the point straight from the definition
Why: The second piece applies exactly at 2.
\[ g(2) = 1 \]
For the limit, use only inputs other than 2
Why: The first piece governs every such input.
\[ g(x) = x + 2\text{ near } 2 \]
Find where those outputs are heading
Why: The linear expression approaches 4.
\[ \lim = 4 \]
Compare
Why: They differ, and both statements are correct.
\[ \lim 4,\text{ value } 1 \]
Figure (svg): The solution to Worked example the value at the point is irrelevant shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 2} g(x) = 4 \ne 1 = g(2) \]
Verify: check that changing the value at 2 changes nothing
Why: Redefine g of 2 to be 100, or leave it undefined entirely: the limit stays 4 in every case, because no input equal to 2 was ever consulted. This is not a curiosity — it is the property that makes limits useful. It also names precisely what continuity will require in Section 2.4: that the limit and the value agree, which here they do not.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 120-121
Trap
\[ g(x) = \begin{cases} x+2 & x \ne 2 \\ 1 & x = 2\end{cases} \]
Substitute the point to get the limit
Why: The student evaluates g at 2 and reports that.
\[ \lim_{x \to 2} g(x) = 1 \quad \text{(wrong)} \]
The outputs near 2 are all near 4, not near 1. A single misplaced point cannot change where the rest of the curve is heading.
\[ \lim_{x \to 2} g(x) = 4, \qquad g(2) = 1 \]
Use only inputs different from the point
Why: The limit is defined in terms of nearby inputs, and the point itself is explicitly excluded.
Substitution happens to give the right answer for most functions you have met, which is exactly why the habit is dangerous. It works precisely when the function is continuous at the point — and Section 2.4 is devoted to saying when that is. Until then, substitution is a guess to be checked, not a method.
Fill the middle
The quotient from the worked example, simplified away from the point.
Fill in the blanks
\frac2___ = x + 1 \;(x \ne 1) \;\Longrightarrow\; \lim____ \frac______ = ___
Why: The simplified expression is defined at 1 and gives 2. The cancellation is legitimate because the two expressions agree at every input the limit actually uses — that is, every input except 1 itself.
Two truths and a lie
All three are about the definition.
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. The quotient x squared minus 1 over x minus 1 is undefined at 1 and has the limit 2 there. Requiring the function to be defined would make the definition useless for the very problems it was invented for, since a difference quotient is always undefined at the point of interest.
Prediction
Commit before reasoning.
Predict first
Why does the definition insist on inputs near a but not equal to a?
Correct: So that a limit can exist where the function is undefined — the case the whole subject needs.
\[ f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}: \quad \text{undefined at } x = a, \text{ always} \]
Why: Every derivative is the limit of a difference quotient, and that quotient is undefined at exactly the point being approached. If the definition required the function to be defined there, no derivative could ever be computed. The exclusion is therefore not a technicality but the feature that makes the definition fit its purpose, and it is also what allows a limit and a value to disagree — which is precisely what continuity in Section 2.4 will rule out.
Section
Section 2
Concept
A limit can be estimated by tabulating values at inputs closing in from both directions, or by reading a graph. Both are evidence rather than proof: a table samples finitely many inputs, and a graph is drawn at finite resolution.
numerical estimation — Tabulating a function at inputs approaching the point from both sides and reading the value the outputs stabilise toward. The estimate is only as good as the inputs chosen and the precision available.
\[ \text{tabulate both sides} \;\Longrightarrow\; \text{compare} \;\Longrightarrow\; \text{report the common trend} \]
Approaching from both sides is not optional care. A one-sided table cannot detect the commonest failure of all, which is that the two sides head for different values.
Figure (svg): A two-sided table of values closing in on four from both directions
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 117-124 — estimating limits numerically and graphically
Picture it
Values closing on the same number from below and from above.
Figure (svg): A two-sided table of values closing in on four from both directions
The columns agree, so the two-sided limit is 4. Had they closed on different values, this table would have shown it immediately and a one-sided table would not.
Worked example
Example 2.6. Both sides, several rows.
\[ \text{Estimate } \lim_{x \to 0} \frac{\sin x}{x} \text{ from a table.} \]
Note that direct substitution fails
Why: Both parts vanish at 0.
\[ \frac{0}{0} \]
Tabulate from the left, in radians
Why: At negative 0.1, negative 0.01, negative 0.001.
\[ 0.99833, 0.99998, 0.9999998 \]
Tabulate from the right
Why: The function is even, so the values match.
\[ 0.99833, 0.99998, 0.9999998 \]
Read the stabilising digits
Why: The values close on 1 from below on both sides.
\[ \lim = 1 \]
Figure (svg): The solution to Worked example a limit estimated from a table shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Verify: check that the units are radians and that the values approach from below
Why: In degrees the same table would give about 0.01745, not 1 — the limit depends entirely on the angle measure, which is the promise Section 1.3 made about why radians matter. The values also approach 1 from below on both sides and never exceed it, consistent with the sine being slightly less than its angle for small positive angles. This limit is the single most important one in the course: Section 3.5 uses it to prove that the derivative of sine is cosine.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 122-123
Fill the middle
Values approaching from below, each row adding a digit.
Fill in the blanks
3.9, \; 3.99, \; 3.999, \; 3.9999 \;\longrightarrow\; 4
Why: The destination is 4, which no row attains. Reading the stabilising digits and extrapolating is what a table is for; reporting the final row confuses an approximation with the limit.
Worked example
Example 2.7. The same method, badly applied.
\[ \text{Evaluate } \lim_{x \to 0} \sin\!\left(\frac{\pi}{x}\right) \text{ using the inputs } 1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}. \]
Evaluate at each chosen input
Why: Pi over one half is 2 pi, and so on: all whole multiples of pi.
\[ \sin(\pi), \sin(2 \pi), \sin(3 \pi),... \]
Note every value
Why: The sine of any whole multiple of pi is zero.
\[ \text{all values are } 0 \]
Read the naive conclusion
Why: The table suggests a limit.
\[ \text{seems to be } 0 \]
Test with different inputs
Why: At 2 over 5 the argument is 5 pi over 2, whose sine is 1.
\[ \text{value } 1,\text{ not near } 0 \]
Conclude
Why: The values do not settle at all.
Figure (svg): A table of values that suggests a limit of zero for a function that has no limit at all
\[ \lim_{x \to 0} \sin\!\left(\tfrac{\pi}{x}\right) \text{ does not exist} \]
Verify: find inputs arbitrarily near 0 giving every value between -1 and 1
Why: Taking x equal to 2 over (4n plus 1) gives the value 1 for every whole n, and these inputs approach 0. Taking x equal to 2 over (4n plus 3) gives negative 1, and these also approach 0. So arbitrarily close to 0 the function takes both extreme values, and it cannot be closing in on any single number. The first table sampled only the zeros, which is exactly the hazard: a table shows what you asked it, not what the function does.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 123-124
Error analysis
A student estimates a limit from a table.
Annotate
On: \( f(1.9) = 3.9, \; f(1.99) = 3.99, \; f(1.999) = 3.999 \;\Longrightarrow\; \lim_{x \to 2} f(x) = 3.999 \)
A limit is never a row of the table. Read which digits have stabilised and extrapolate: here three 9s after the point in a value below 4 says the destination is 4, and the second column confirms it from the other side.
Matching
What each pattern of values supports.
Match the pairs
Why: Only the first supports a limit. The other three are the three failure modes, and distinguishing them matters: the second still has useful one-sided information, while the third and fourth have none.
Two truths and a lie
All three are about estimating limits.
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. No finite table establishes a limit — the sine of pi over x gives exactly 0 at infinitely many inputs approaching 0, and has no limit at all. A table can be made to agree to any number of digits and still be wrong. Tables build intuition and suggest answers; only algebra or the precise definition of Section 2.5 establishes one.
Prediction
Commit before reasoning.
Predict first
What can a two-sided table detect that a one-sided one cannot?
Correct: That the two sides head for different values.
\[ \lim_{x \to 1^-} = 1, \quad \lim_{x \to 1^+} = 2 \;\Longrightarrow\; \lim_{x \to 1} \text{ does not exist} \]
Why: A jump discontinuity looks perfectly well behaved from one side: the values close on a number, the digits stabilise, everything appears settled. Only comparing with the other side reveals that they settle on different numbers and the two-sided limit does not exist. Since jumps are the commonest failure in practice — every piecewise model has candidates for one — checking both sides is the single highest-value habit in this section.
Section
Section 3
Concept
The left-hand limit uses only inputs less than the point; the right-hand limit only inputs greater. Each can exist on its own, and the two-sided limit exists exactly when both exist and are equal.
one-sided limits — The left-hand limit is the value approached using only inputs below the point, written with a minus superscript; the right-hand limit uses only inputs above, written with a plus. The two-sided limit exists precisely when both exist and agree.
\[ \lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L = \lim_{x \to a^+} f(x) \]
The theorem is an if-and-only-if, so it works in both directions. It is how a two-sided limit is computed for a piecewise function, and it is how a limit is proved not to exist.
Figure (svg): A piecewise function with different one-sided limits, each approach traced separately
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 126-131 — one-sided limits
Picture it
A piecewise function whose sides disagree at the seam.
Figure (svg): A piecewise function with different one-sided limits, each approach traced separately
Both one-sided limits exist and are perfectly respectable numbers. They simply differ, so the two-sided limit does not exist — and saying exactly that is far more informative than saying the limit fails.
Worked example
Example 2.10. Each side uses its own rule.
\[ \text{For } f(x) = \begin{cases} x^2 & x < 1 \\ 3 - x & x > 1 \end{cases} \text{ find both one-sided limits at } 1. \]
For the left-hand limit, use inputs below 1
Why: The first rule governs them.
\[ \lim\text{ from left of } x ^{2} \]
Evaluate
Why: The squaring function approaches 1.
\[ \text{left } \lim = 1 \]
For the right-hand limit, use inputs above 1
Why: The second rule governs them.
\[ \lim\text{ from right of } 3 - x \]
Evaluate
Why: Three minus 1.
\[ \text{right } \lim = 2 \]
Compare
Why: They differ, so the theorem's condition fails.
Figure (svg): The solution to Worked example one-sided limits of a piecewise function shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 1^-} f(x) = 1, \quad \lim_{x \to 1^+} f(x) = 2 \]
Verify: confirm the value at 1 plays no part
Why: The definition given assigns no value at 1 at all, and both one-sided limits were computed regardless — because each uses only inputs strictly on its own side. Defining f of 1 to be 1, or 2, or 17 would change none of these three answers. The size of the jump, 2 minus 1, is 1, and Section 2.4 will call this a jump discontinuity precisely because both one-sided limits exist and differ.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 128-129
Matching
The superscript names the direction of approach.
Match the pairs
Why: Four statements about the same point, using four different sets of inputs — and the last uses a set the other three deliberately exclude. Keeping them apart is what makes the definition of continuity in Section 2.4 say something rather than being circular.
Worked example
Checkpoint 2.10. Both sides agree, so the limit exists.
\[ \text{For } g(x) = \begin{cases} 2x + 1 & x \le 3 \\ x + 4 & x > 3 \end{cases} \text{ find } \lim_{x \to 3} g(x). \]
Compute the left-hand limit
Why: The first rule applies below 3.
\[ 2(3) + 1 = 7 \]
Compute the right-hand limit
Why: The second rule applies above 3.
\[ 3 + 4 = 7 \]
Compare
Why: They agree.
\[ \text{both equal } 7 \]
Apply the theorem
Why: Existence and equality of both sides give the two-sided limit.
\[ \lim = 7 \]
Figure (svg): The solution to Worked example using the theorem in the other direction shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 3} g(x) = 7 \]
Verify: check the value at the point separately
Why: The first piece owns 3, since its condition carries the equals sign, so g of 3 is 7 as well. Here the limit and the value agree, which is the definition of continuity that Section 2.4 will state — so this function has no break at 3, unlike the previous example. It is worth noticing that these are two separate checks: the limit came from the two one-sided computations, and the value from reading which piece owns the boundary.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 129-130
Trap
\[ f(x) = \begin{cases} x^2 & x < 1 \\ 3-x & x > 1\end{cases} \]
Compute the left-hand limit with the second rule
Why: The student picks the piece by its position on the page rather than by its condition.
\[ \lim_{x \to 1^-} f(x) = 3 - 1 = 2 \quad \text{(wrong)} \]
Inputs approaching 1 from below are all less than 1, and the second rule does not apply to any of them.
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1 \]
Match the direction of approach to the condition on the piece
Why: A left-hand limit uses inputs below the point, so it uses the rule stated for inputs below the point.
The mnemonic that removes the error entirely: the minus superscript means coming from the smaller numbers, so use the rule for smaller numbers. Writing the condition beside the limit before evaluating anything takes two seconds and makes the choice mechanical.
Fill the middle
The left-hand limit of the piecewise function from the worked example.
Fill in the blanks
\lim_x^2 f(x) = \lim____ ___ = 1
Why: Inputs approaching from the left are all below 1, so the rule stated for inputs below 1 applies. Using the other rule would have given 2, which is the right-hand limit and answers a different question.
Sorting
Apply the theorem: both must exist and agree.
Sort into buckets
Sort each pair of one-sided limits.
The last case is worth noticing: one side behaving perfectly is not enough. The theorem is a conjunction, and a single failing side defeats it, which is why both must always be checked.
Prediction
Commit before reasoning.
Predict first
Is it possible for a left-hand limit to exist while the right-hand limit does not?
Correct: Yes. The two use disjoint sets of inputs and are entirely independent.
\[ \lim_{x \to 0^-} f = 0 \quad \text{while} \quad \lim_{x \to 0^+} f \text{ does not exist} \]
Why: Take the function equal to x for negative inputs and the sine of one over x for positive ones. From the left it approaches 0 cleanly; from the right it oscillates forever and has no limit. Nothing links the two sides, because no input is used by both computations. This independence is exactly why the theorem has to demand both conditions, and why checking only one side proves nothing about the two-sided limit.
Section
Section 4
Concept
A two-sided limit fails to exist in exactly three ways: the one-sided limits exist but differ, the values grow without bound, or the values oscillate without settling. Naming which one applies is more useful than merely reporting failure.
failure of a limit — A limit fails to exist when no single number is approached. The three mechanisms are a jump, where the one-sided limits differ; unbounded growth; and endless oscillation.
\[ \text{jump} \quad \text{blow-up} \quad \text{oscillation} \]
The three are genuinely different in what information survives. A jump leaves both one-sided limits intact; a blow-up leaves a description in terms of infinity; an oscillation leaves nothing at all.
Figure (svg): The three ways a limit can fail: a jump, a blow-up, and an endless oscillation
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 124-133 — when limits fail to exist
Picture it
One picture each.
Figure (svg): The three ways a limit can fail: a jump, a blow-up, and an endless oscillation
The first still has useful one-sided limits, the second has a vertical asymptote to describe, and the third has nothing to say beyond that the values never settle. Recognising which you are in tells you what can still be salvaged.
Worked example
Example 2.8. Three functions, three diagnoses.
\[ \text{Say why each fails at } 0: \; \frac{|x|}{x}, \; \frac{1}{x^2}, \; \sin\!\left(\frac{1}{x}\right). \]
Examine the first from both sides
Why: The absolute value over x is negative 1 below and 1 above.
\[ \text{left } -1,\text{ right } 1 \]
Diagnose it
Why: Both one-sided limits exist and differ.
Examine the second
Why: The square keeps the denominator positive and tiny.
Diagnose it
Why: Unbounded above.
Examine the third
Why: The argument races through every value as x nears 0.
\[ \text{takes } 1\text{ and } -1\text{ arbitrarily near } 0 \]
Diagnose it
Why: It never settles.
Figure (svg): The solution to Worked example identifying the failure mode shown as a ladder of expressions, one row per legal move
\[ \text{jump}, \quad \text{blow-up}, \quad \text{oscillation} \]
Verify: check what survives in each case
Why: For the first, the one-sided limits negative 1 and 1 are perfectly good numbers and the jump has size 2. For the second, the statement that the limit is infinity is a genuine description that locates a vertical asymptote. For the third, nothing survives — there is no one-sided limit on either side and no asymptote. That the three failures leave different amounts of information is precisely why naming them is worth the trouble.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 125-126
Sorting
Diagnose each, at the point named.
Sort into buckets
Sort each function.
The two blow-ups differ in an important way: one over x squared goes to positive infinity from both sides, so its limit can be described as infinity, while one over x goes to opposite infinities and cannot even be described that way. Both have a vertical asymptote, but only one has a limit statement.
Worked example
Checkpoint 2.8. Jumps are not pathological; they are common.
\[ \text{Postage is } \$1 \text{ up to } 1 \text{ oz and } \$1.20 \text{ above. Examine the limit at } 1 \text{ oz.} \]
Compute the limit from below
Why: Weights just under an ounce all cost a dollar.
\[ \text{left } \lim = 1.00 \]
Compute the limit from above
Why: Weights just over cost a dollar twenty.
\[ \text{right } \lim = 1.20 \]
Compare
Why: They differ by twenty cents.
Interpret the jump
Why: The price changes abruptly at a threshold.
\[ a\text{ genuine jump of } \$ 0.20 \]
Figure (svg): The solution to Worked example a jump from a real model shown as a ladder of expressions, one row per legal move
\[ \lim_{w \to 1^-} = 1.00 \ne 1.20 = \lim_{w \to 1^+} \]
Verify: ask whether the jump is an artefact or the point
Why: It is the point. The pricing rule genuinely changes at one ounce, and the discontinuity is a faithful description of that policy rather than a defect in the model. Note that the value AT one ounce is a separate question settled by the wording of the rule — up to and including one ounce means the value is 1.00, matching the left-hand limit only. Real-world piecewise rules produce jumps routinely, which is why the vocabulary is worth having.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 126-127
Error analysis
A student evaluates a limit using a calculator table.
Annotate
On: \( \sin\!\left(\tfrac{\pi}{x}\right) \text{ at } x = 1, \tfrac{1}{2}, \tfrac{1}{3} \text{ all give } 0 \;\Longrightarrow\; \lim_{x \to 0} = 0 \)
A table reports what you asked it, not what the function does. When a function oscillates, a regularly spaced set of inputs can land on the same phase every time and produce a table of perfect, meaningless consistency.
Fill the middle
The absolute value of x divided by x, examined on each side of zero.
Fill in the blanks
\lim_1 \frac______ = -1, \qquad \lim____ \frac______ = ___
Why: For positive inputs the absolute value of x is x, so the quotient is 1. For negative inputs it is negative x, giving negative 1. Both sides exist and differ, which is a jump of size 2.
Two truths and a lie
All three are about failures.
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 conflates two independent things. The absolute value of x over x is undefined at 0, but the postage function is perfectly well defined at one ounce and still has no limit there. Whether the function has a value at the point and whether it has a limit there are separate questions — which is exactly why continuity in Section 2.4 needs to demand both plus their agreement.
Prediction
Commit before reasoning.
Predict first
Which failure mode leaves you able to say the most about the function near the point?
Correct: The jump — both one-sided limits survive as ordinary numbers.
\[ \text{jump: } L^- \text{ and } L^+ \text{ both exist}; \quad \text{oscillation: neither does} \]
Why: After a jump you can state exactly what the function approaches from each side and how large the step is, which is often all the application needs. A blow-up leaves only the asymptote and a direction. An oscillation leaves nothing: the function is bounded, which sounds reassuring, but no approach value exists on either side. Boundedness is not the same as convergence, and conflating them is a common source of error.
Section
Section 5
Concept
When the outputs grow beyond every bound as the input approaches a point, we write that the limit is infinity. This is a description of a particular failure, not a claim that the limit exists — infinity is not a number the function is approaching.
infinite limit and vertical asymptote — If the outputs increase beyond every bound as the input approaches a from either side, the limit is said to be infinity. If either one-sided limit is infinite, the line through a parallel to the vertical axis is a vertical asymptote.
\[ \lim_{x \to 0} \frac{1}{x^2} = \infty, \qquad \text{vertical asymptote at } x = 0 \]
One asymptote condition is enough. The reciprocal function has one-sided limits of negative and positive infinity, so no limit at all can be written, yet the vertical asymptote is still there — because only one side needs to blow up.
Figure (svg): Two functions with vertical asymptotes: one going to positive infinity on both sides, one with opposite signs
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 133-139 — infinite limits and vertical asymptotes
Picture it
One function blowing up the same way on both sides, one differently.
Figure (svg): Two functions with vertical asymptotes: one going to positive infinity on both sides, one with opposite signs
Both have a vertical asymptote at the origin. Only the first supports a two-sided infinite limit statement; for the second, the sides disagree in sign and only one-sided statements are available.
Worked example
Example 2.11. Determine the behaviour on each side.
\[ \text{Evaluate } \lim_{x \to 0} \frac{1}{x^2} \text{ and } \lim_{x \to 0} \frac{1}{x}. \]
For the first, note the denominator's sign
Why: A square is positive for every non-zero input.
Conclude for the first
Why: A fixed numerator over a tiny positive number is huge and positive, on both sides.
\[ \lim = \infty \]
For the second, take each side separately
Why: The denominator's sign now follows the input's.
State both one-sided limits
Why: They blow up in opposite directions.
Conclude for the second
Why: No single description covers both sides.
Figure (svg): The solution to Worked example an infinite limit shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0}\frac{1}{x^2} = \infty, \qquad \lim_{x \to 0}\frac{1}{x} \text{ does not exist} \]
Verify: check that both still have a vertical asymptote
Why: Both functions have a vertical asymptote at the origin, because an asymptote needs only ONE side to blow up. So the two questions are different: whether an infinite limit can be written asks about both sides agreeing, while whether an asymptote exists asks about either side. Reporting that the reciprocal has no asymptote because its limit does not exist would be a real error.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 134-136
Matching
Check each side separately.
Match the pairs
Why: The last is the important reminder: a zero denominator does not guarantee an asymptote. Here the numerator vanishes at the same rate, the quotient has the finite limit 1, and there is no asymptote at all.
Worked example
Checkpoint 2.11. Where the denominator vanishes, usually.
\[ \text{Find the vertical asymptotes of } f(x) = \frac{x+1}{x^2 - 4}. \]
Factor the denominator
Why: A difference of squares.
\[ (x - 2) (x + 2) \]
Find where it vanishes
Why: Each factor set to zero.
\[ x = 2\text{ and } x = -2 \]
Check the numerator does not vanish there
Why: At 2 it is 3; at negative 2 it is negative 1.
Conclude
Why: Both are genuine asymptotes.
\[ \text{asymptotes at } x = 2\text{ and } x = -2 \]
Figure (svg): The solution to Worked example locating asymptotes shown as a ladder of expressions, one row per legal move
\[ x = 2 \text{ and } x = -2 \]
Verify: contrast with a case where the numerator does vanish
Why: Had the numerator been x minus 2, the factor would have cancelled and x equal to 2 would be a HOLE rather than an asymptote — the function would have a perfectly finite limit there. So a zero denominator is not by itself enough; what matters is whether the numerator vanishes too, and whether the cancellation removes the problem entirely. Checking the numerator at each candidate is what separates the two cases.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 137-138
Trap
\[ \lim_{x \to 0} \frac{1}{x^2} = \infty \]
Conclude that the limit exists and equals infinity
Why: The student reads the equals sign as an ordinary one.
\[ \text{so } \tfrac{1}{x^2} \text{ 'converges to' } \infty \quad \text{(wrong reading)} \]
Infinity is not a number, so nothing can get arbitrarily close to it. The statement is a description of unbounded growth, not a convergence claim.
\[ \lim_{x \to 0}\frac{1}{x^2} = \infty \;\text{ means: the values exceed every bound} \]
Read the notation as shorthand for a specific failure
Why: It says that for any bound, however large, all inputs close enough to 0 give outputs beyond it.
The distinction has teeth. Because infinity is not a number, the limit laws of Section 2.3 do not apply to these statements — subtracting one infinite limit from another is not legal and gives no answer, which is why expressions of that form are called indeterminate and need the separate treatment of Section 4.8.
Fill the middle
The rational function from the worked example, with its denominator factored.
Fill in the blanks
x^2 - 4 = (x-2)(x+2) = 0 \;\Longrightarrow\; x = 2 \text-2 x = ___
Why: Both factors give candidate asymptotes, and both are genuine because the numerator does not vanish at either. Had the numerator shared a factor, that candidate would have been a hole instead.
Sorting
A vanishing denominator is only a candidate; check the numerator.
Sort into buckets
Sort each, at the input where the denominator vanishes.
The last case has no common factor to cancel and is still a hole, because the numerator approaches zero just as fast as the denominator. That the sine limit is 1 rather than 0 or infinity cannot be seen by factoring at all, which is why Section 2.3 needs the Squeeze Theorem to settle it.
Prediction
Commit before reasoning.
Predict first
The denominator of a rational function vanishes at x = 2. Must there be a vertical asymptote there?
Correct: No. If the numerator vanishes too, cancellation may leave a finite limit.
\[ \frac{x^2-4}{x-2} = x+2 \;(x \ne 2) \;\Longrightarrow\; \lim_{x \to 2} = 4 \]
Why: For the quotient of x squared minus 4 by x minus 2, the denominator vanishes at 2 but so does the numerator, and cancelling leaves x plus 2, which has the perfectly finite limit 4. The graph has a hole at that input, not an asymptote. This is exactly the situation of Section 2.1's secant slopes: zero over zero is a prompt to simplify, and only after simplifying can you tell which case you are in.
Comparison
Fill the blanks. Four different questions about the same point.
Comparison matrix
| Question | Which inputs it uses | Can it exist when the others do not? |
|---|---|---|
| Left-hand limit | inputs strictly below a | yes, independently of the right |
| Right-hand limit | inputs strictly above a | yes, independently of the left |
| Two-sided limit | inputs on both sides, never a itself | only if both one-sided limits exist and agree |
| The value f(a) | the input a alone | yes, entirely independent of all three limits |
The last row is the one worth dwelling on. The value and the limit are computed from disjoint sets of inputs, so neither constrains the other — and continuity, in Section 2.4, is precisely the demand that they agree anyway.
Pattern
Given a limit to evaluate at a point.
Step one is not laziness. For every polynomial, and for every rational function away from its denominator's zeros, substitution is exactly right — and Section 2.3 will prove it rather than leaving it as a hopeful habit.
Stewart, Calculus: Early Transcendentals 8e, §2.2 The Limit of a Function §2.2, pp. 83-94
Check
Near, not at.
Check your understanding
If g(x) = x + 2 for every x except 2, where g(2) = 1, what is the limit of g as x approaches 2?
Answer: A
Why: The limit uses only inputs other than 2, where the rule is x + 2, so it approaches 4.
Check
One-sided limits. Match rule to side.
Check your understanding
For f equal to x^2 below 1 and 3 - x above 1, what is the limit from the left at 1?
Answer: A
Why: Inputs below 1 use the squaring rule, which approaches 1.
Check
Asymptote or hole? Check the numerator too.
Check your understanding
Does f(x) = (x^2 - 4)/(x - 2) have a vertical asymptote at x = 2?
Answer: A
Why: The numerator factors as (x-2)(x+2), so the quotient is x + 2 away from 2, with limit 4.
Real world
A streaming service charges nothing for the first 10 hours a month, then 50 cents per hour beyond that. Separately, a chemical reaction's rate is measured as the concentration of a reagent approaches zero.
Discussion prompt
For the billing rule, find both one-sided limits at 10 hours and say what the discontinuity means. For the reaction, explain why the measured rate at exactly zero concentration cannot be observed and what is reported instead.
Hint: One is a jump; the other is a limit at a point the experiment cannot reach.
Answer:
The billing rule. Just below 10 hours the charge is 0; just above it is still near 0, since the first hour beyond costs only pennies. So both one-sided limits are 0 and the charge function is actually continuous at 10 — the RATE jumps, but the total does not.
\[ \lim_{h \to 10^-} C(h) = 0 = \lim_{h \to 10^+} C(h) \]
The jump is in the derivative, not the function: the slope goes from 0 to 0.5 abruptly. That is a corner, and Section 3.2 will show it is exactly where a derivative fails to exist even though the function is perfectly continuous.
The reaction. The rate at exactly zero concentration cannot be measured, because with no reagent there is no reaction to observe — the experiment is undefined at that point, just as the difference quotient is undefined at the point of tangency.
What chemists report is the limiting rate: measure at concentrations of 0.1, 0.01, 0.001 and read where the values are heading. This is a limit in exactly the sense defined here, and it carries the same warning — the measurements are evidence, the trend must be read rather than the last row, and instrument noise grows as the concentration shrinks, so the table cannot be pushed indefinitely.
\[ \text{rate}_0 = \lim_{c \to 0^+} \text{rate}(c) \quad \text{- a one-sided limit, since } c < 0 \text{ is meaningless} \]
Commit first
Answer, then rate your confidence honestly.
Predict first
What does writing that a limit equals infinity assert?
Correct: That the outputs exceed every bound — a description of how the limit fails.
\[ \lim_{x \to a} f(x) = \infty: \quad \text{for every } M, \; f(x) > M \text{ for } x \text{ near } a \]
Why: Infinity is not a number, so nothing can get arbitrarily close to it and no convergence is being claimed. The notation is shorthand for unbounded growth, and the practical consequence is that the limit laws of Section 2.3 do not apply to it — you cannot subtract one infinite limit from another. Whether the function is defined at the point is a separate question, and an unbounded vertical behaviour gives a VERTICAL asymptote, not a horizontal one.
Explain it
They cannot accept that a function undefined at a point still has a limit there. It feels to them like answering a question about nothing.
Discussion prompt
In four sentences or fewer, explain why a limit can exist where the function does not.
Hint: Ask them what inputs the limit actually uses.
Answer:
Ask which inputs the limit looks at. The answer is: every input near the point except the point itself — that exclusion is written into the definition. So a hole at the point removes an input the limit was never going to consult.
Think of driving toward a town along a straight road. You can say with total confidence where the road is heading without knowing anything about the town, or even whether it exists. The limit asks about the road, not the destination — and that is why every derivative in Chapter 3, which is a quotient undefined at exactly the point of interest, has an answer at all.
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 limit versus value, ask which inputs the question uses — everything near, or the point alone. For one-sided limits, write the condition beside the limit before evaluating: the minus superscript means smaller inputs. For failures, check both sides first, then whether the values are bounded. For asymptote versus hole, always check whether the numerator vanishes too, and factor before deciding. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Across the top of the page draw three small graphs of functions that all have limit 4 at the input 2: one continuous there, one with a hole, one with the value placed at 1. Under them write the single sentence explaining why all three have the same limit. In the middle, draw the three failure modes side by side — a jump, a blow-up and an oscillation — and beside each write which one-sided limits exist and what information survives. Below that, take the piecewise function that is x squared below 1 and 3 minus x above, compute both one-sided limits showing which rule you used for each, and state whether the two-sided limit exists. Then take the quotient of x squared minus 4 by x minus 2, show the cancellation, state the limit at 2, and write one sentence saying why this is a hole and not an asymptote. In a margin, write what it means to say a limit is infinity, and one sentence on why the limit laws will not apply to it.
If your three top graphs look different anywhere other than at the single input 2, redraw them: the whole point is that they are identical everywhere else, which is what makes their sharing a limit inevitable rather than surprising.
Recap
Five things, and the first is one sentence that the whole of Chapters 3 and 5 depend on.
| If you see | Then |
|---|---|
| A hole at the point | The limit may still exist |
| A value placed off the curve | It cannot change the limit |
| A piecewise rule | Compute both one-sided limits |
| The sides disagreeing | No two-sided limit: a jump |
| A vanishing denominator only | A vertical asymptote |
| Both parts vanishing | Factor and cancel: probably a hole |
| Values oscillating | No limit, and no one-sided limits either |
Section 2.3 stops estimating. It gives the limit laws that let most limits be evaluated exactly and in one line, plus the two techniques — algebraic simplification and the Squeeze Theorem — that handle the cases the laws cannot reach on their own.
OpenStax Calculus Volume 1, §2.2 The Limit of a Function §2.2, pp. 116-139 — everything on these slides traces back here
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