Week 1 gave you functions to work with. This week asks what it means for a function's output to get close to a number as the input gets close to another number — the single idea that the derivative (Week 4), the integral (Week 7), and the whole theory of series (Weeks 14–16) are all built from.
2.1 Two Problems, One Idea
Two questions drive this week, and Thomas' Calculus opens with them for a reason: both turn out to be the same question in disguise.
- The tangent problem. Given a curve and a point on it, what is the slope of the line that just touches the curve there? Ordinary slope needs two points; a tangent line is defined by one.
- The instantaneous rate problem. Given a quantity that changes over time, how fast is it changing at one exact instant? Ordinary rate of change needs an interval; an instant is a single point in time.
Both problems ask for information at a single point using a formula that fundamentally needs two. The resolution is the same in both cases: compute the two-point version on a small interval, then ask what happens as that interval shrinks to nothing. That question — "what happens as something shrinks to nothing" — is a limit.
2.2 Average Rate of Change and Secant Lines
For a function y = f(x), the average rate of change over the interval [x1, x2] is
Geometrically, this is the slope of the secant line — the line through the two points (x1, f(x1)) and (x2, f(x2)) on the graph. It is a perfectly good answer to "how fast, on average, did y change between these two points" — it says nothing about the rate at any single instant inside the interval.
Example 2.1 — Average rate of change
A ball's height is h(t) = 20t − 5t2 metres, t seconds after release. Find its average velocity over [1, 3].
Step 1 — Evaluate the function at the endpoints
h(1) = 20(1) − 5(1) = 15; h(3) = 20(3) − 5(9) = 60 − 45 = 15
Step 2 — Apply Eq. 2.1
Average velocity = h(3) − h(1)3 − 1 = 15 − 152 = 0
The ball has the same height at t = 1 and t = 3, so its net displacement over that interval is zero — even though it was moving (up, then down) throughout.
2.3 The Tangent Line Problem
To get the slope at a single point a rather than between two points, fix x1 = a and let the second point be a + h for a small number h. The secant slope through these two points is
This is exactly Eq. 2.1 with x2 − x1 renamed h. As h shrinks toward 0, the second point slides along the curve back toward a, and the secant line rotates toward the tangent line at a. The slope of the tangent line is therefore
Equation 2.3 is exactly the definition of the derivative that Week 4 will introduce formally and name f′(a). You are seeing it now because it is the reason limits matter: the whole differentiation chapter is this one limit, evaluated for every kind of function in turn.
2.4 An Informal Definition of Limit
We say
— read "the limit of f(x) as x approaches a equals L" — if the values of f(x) get arbitrarily close to L as x gets close to a from either side, without ever needing x = a itself. Three points are easy to miss the first time:
- The limit describes behaviour near a, not at a. It exists even if f(a) is undefined, and it can differ from f(a) when both exist.
- "Arbitrarily close" must hold from both directions — approaching along x < a and along x > a must give the same target value L. Week 3 names these one-sided limits explicitly.
- A limit is a single proposed number L, checked against the function's behaviour — it is not computed by "plugging in a slightly different x," though that is a fine way to guess it before proving it.
2.5 When a Limit Fails to Exist
Not every function has a limit at every point. Three common ways a limit can fail to exist:
| Behaviour | Example near x = 0 |
|---|---|
| Function approaches different values from each side | f(x) = x/|x|: −1 from the left, +1 from the right |
| Function grows without bound | f(x) = 1/x2: increases without bound as x → 0 |
| Function oscillates infinitely often | f(x) = sin(1/x): keeps oscillating between −1 and 1, never settling |
The first case is resolved precisely once Week 3 defines one-sided limits: the left-hand and right-hand limits both exist here, they are just unequal, so the two-sided limit does not exist. The other two cases need the language of infinite limits, also in Week 3.
2.6 The Limit Laws
If limx→a f(x) = L and limx→a g(x) = M, both limits existing, then:
Two limits used constantly as starting points, both provable directly from the definition:
2.7 Limits by Direct Substitution
Applying the limit laws repeatedly to a polynomial or rational function shows that its limit at a can be found simply by substituting x = a, provided the result is not an undefined expression:
Example 2.2 — Direct substitution
Evaluate limx→2 (3x2 − 5x + 1).
Step 1 — Check that x = 2 is in the domain
Polynomials are defined everywhere, so direct substitution applies.
Step 2 — Substitute
3(2)2 − 5(2) + 1 = 12 − 10 + 1
= 3
2.8 Removing Zero-Denominator Indeterminate Forms
Direct substitution fails when it produces the meaningless expression 0/0, called an indeterminate form. This does not mean the limit fails to exist — it means substitution cannot answer the question, and an algebraic rewrite is needed first. Two techniques handle almost every case in this course: factoring and rationalizing.
Example 2.3 — Removing a 0/0 form by factoring
Evaluate limx→3 x2 − 9x − 3.
Step 1 — Try direct substitution
(9 − 9)/(3 − 3) = 0/0 — indeterminate, so substitution alone cannot answer this.
Step 2 — Factor and cancel the common factor
x2 − 9x − 3 = (x−3)(x+3)x − 3 = x + 3, valid for every x ≠ 3
The cancellation is legal precisely because a limit as x → 3 never looks at x = 3 itself (§2.4), so the two sides of the equation agree everywhere the limit actually examines.
Step 3 — Substitute into the simplified expression
limx→3 (x + 3) = 6
Example 2.4 — Removing a 0/0 form by rationalizing
Evaluate limx→0 √(x+4) − 2x.
Step 1 — Try direct substitution
(√4 − 2)/0 = 0/0 — indeterminate.
Step 2 — Multiply by the conjugate of the numerator
√(x+4) − 2x · √(x+4) + 2√(x+4) + 2 = (x+4) − 4x(√(x+4) + 2) = xx(√(x+4) + 2)
Step 3 — Cancel x and substitute
1√(x+4) + 2, then at x = 0: 1/(√4 + 2) = 1/4
limx→0 √(x+4) − 2x = 1/4
2.9 The Squeeze Theorem
Sometimes neither factoring nor rationalizing is available — the function that needs bounding is transcendental. If g(x) ≤ f(x) ≤ h(x) for all x near a (except possibly at a itself), and
then limx→a f(x) = L as well. If f is squeezed between two functions that both converge to L, it has no room to converge to anything else.
Example 2.5 — Applying the Squeeze Theorem
Evaluate limx→0 x2 sin(1/x).
Step 1 — Bound the oscillating factor
For every x ≠ 0, −1 ≤ sin(1/x) ≤ 1, so −x2 ≤ x2 sin(1/x) ≤ x2.
Step 2 — Take the limit of both bounding functions
limx→0 (−x2) = 0 and limx→0 x2 = 0 (Eq. 2.6).
By the Squeeze Theorem, limx→0 x2 sin(1/x) = 0, even though sin(1/x) itself has no limit at 0 (§2.5).
2.10 The Precise Definition of a Limit (Not Examinable)
Per the course schedule, §2.3 (the epsilon–delta definition) is not examinable this term. It is included because it is the rigorous statement behind everything in §2.4–2.9, and reading it once makes the informal language ("arbitrarily close") mean something concrete.
The phrase "f(x) gets arbitrarily close to L" is made precise by requiring that f(x) can be forced within any tolerance ε > 0 of L, simply by keeping x close enough — within some distance δ — to a:
0 < |x − a| < δ ⇒ |f(x) − L| < ε (2.8)
Both absolute values are the distance reading from Week 1 §1.3: |x−a| is how far x is from a, and |f(x)−L| is how far the output is from the claimed limit. The definition says: you name how close to L you want the output (ε), and the function's behaviour must let someone find a matching input-tolerance (δ) that guarantees it — no matter how small a tolerance you demand.
2.11 Worked Capstone Example
Example 2.6 — Combining substitution, factoring, and the limit laws
Evaluate limx→1 x3 − 1x2 − 1.
Step 1 — Try direct substitution (§2.7)
(1−1)/(1−1) = 0/0 — indeterminate.
Step 2 — Factor numerator and denominator (§2.8)
x3 − 1 = (x−1)(x2+x+1); x2 − 1 = (x−1)(x+1)
Step 3 — Cancel the common factor, valid since x ≠ 1 near the limit
(x−1)(x2+x+1)(x−1)(x+1) = x2+x+1x+1
Step 4 — Apply the quotient law and substitute (§2.6)
1+1+11+1 = 32
limx→1 x3 − 1x2 − 1 = 3/2
2.12 Nomenclature
| Symbol | Meaning |
|---|---|
| Δy/Δx | Average rate of change; slope of a secant line |
| msec | Secant line slope |
| limx→a f(x) | The limit of f(x) as x approaches a |
| L | The proposed value of a limit |
| DNE | "Does not exist" — shorthand for a limit that fails to exist |
| ε, δ | Output tolerance and matching input tolerance in the precise definition of a limit |
2.13 Problems
- Average rate of change. For f(x) = x2 + 1, find the average rate of change over [2, 5].
- Secant to tangent. For f(x) = x2, write the secant slope formula (Eq. 2.2) at a = 3, simplify it as far as possible without letting h → 0, then state what it approaches as h → 0.
- Reading a limit statement. Explain, in your own words, why limx→2 f(x) can equal 5 even if f(2) = 7, or if f(2) is undefined.
- Limit that fails to exist. Sketch f(x) = x/|x| and explain why limx→0 f(x) does not exist.
- Direct substitution. Evaluate limx→−1 (2x3 − x + 4).
- Factoring. Evaluate limx→2 (x2−4)/(x−2).
- Factoring, harder. Evaluate limx→−3 (x2+x−6)/(x+3).
- Rationalizing. Evaluate limx→0 (√(x+9) − 3)/x.
- Rationalizing, harder. Evaluate limh→0 (√(1+h) − 1)/h.
- Squeeze Theorem. Evaluate limx→0 x4 cos(1/x2), stating the two bounding functions you use.
- Combined. Evaluate limx→4 (x−4)/(√x−2), and state which technique from §2.8 you used.
- ε–δ, not examinable. For f(x) = 2x+1 and the claim limx→3 f(x) = 7, find a value of δ that works for ε = 0.1.
- Tangent slope from the limit definition. Using Eq. 2.3, find the slope of the tangent line to f(x) = x2 + 1 at a = 3.
2.14 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §2.1 (rates of change and tangent lines), §2.2 (limit of a function and limit laws), §2.3 (the precise definition of a limit, not examinable this term).
- Pearson MyLab Math — the Week 2 homework set is assigned directly from the sections above; see the Calendar for the due date.