Week 2 defined the limit and gave you tools to compute it. This week closes two gaps that came up along the way: what to do when a function behaves differently on each side of a point, and what to do when a function or its input grows without bound.
3.1 One-Sided Limits
The right-hand limit of f at a looks only at values of x greater than a:
meaning f(x) approaches L as x approaches a through values x > a only. The left-hand limit is defined the same way restricted to x < a:
All the limit laws of Week 2 §2.6 apply equally to one-sided limits — nothing about sums, products or quotients of limits depended on approaching from both sides.
3.2 The Two-Sided Limit as an Agreement of Both Sides
This finally makes precise the "different values from each side" failure raised in Week 2 §2.5. For f(x) = x/|x|:
Example 3.1 — Confirming a limit does not exist, formally
Show that limx→0 x/|x| does not exist.
Step 1 — Right-hand limit
For x > 0, |x| = x, so f(x) = x/x = 1. limx→0+ f(x) = 1
Step 2 — Left-hand limit
For x < 0, |x| = −x, so f(x) = x/(−x) = −1. limx→0− f(x) = −1
Step 3 — Compare
1 ≠ −1, so by Eq. 3.3 the two-sided limit does not exist (DNE).
3.3 Continuity at a Point
f is continuous at x = a when three separate conditions all hold:
| # | Condition | What it rules out |
|---|---|---|
| 1 | f(a) is defined | A hole with nothing plotted there |
| 2 | limx→a f(x) exists | A jump between two different one-sided values |
| 3 | limx→a f(x) = f(a) | A hole with a point plotted somewhere else instead |
All three must hold; each rules out a different way a graph can break at a. This is exactly why Example 1.2 in Week 1 (the piecewise function joining smoothly at x=1) looked unbroken: all three conditions happened to hold there.
3.4 Continuity on an Interval
f is continuous on an open interval if it is continuous at every point in it. On a closed interval [a,b], the endpoints only need one-sided continuity: right-continuous at a, left-continuous at b. A function continuous everywhere on its domain (true of every polynomial, rational function away from its zeros of the denominator, and every trigonometric function on its domain) needs no case-by-case checking — direct substitution (Week 2 §2.7) already told you this.
3.5 Types of Discontinuity
| Type | What fails | Fixable? |
|---|---|---|
| Removable | The limit exists but either f(a) is undefined or does not match it (conditions 1 or 3) | Yes — redefine f(a) to equal the limit |
| Jump | Both one-sided limits exist but disagree (Eq. 3.3 fails) | No |
| Infinite | One or both one-sided limits are ±∞ (§3.7) | No |
| Oscillating | The function oscillates infinitely often near a, so no one-sided limit exists | No |
Example 3.2 — Identifying and removing a discontinuity
Let g(x) = (x2−9)/(x−3) for x ≠ 3, undefined at x = 3. Classify the discontinuity at x = 3 and repair it if possible.
Step 1 — Check which condition fails
g(3) is undefined, so condition 1 fails immediately.
Step 2 — Check whether the limit still exists (Week 2, Example 2.3)
limx→3 g(x) = 6 (factor and cancel, exactly as before).
Step 3 — Classify and repair
The limit exists but g(3) does not: a removable discontinuity. Define G(x) = g(x) for x ≠ 3 and G(3) = 6; G is now continuous everywhere.
3.6 The Intermediate Value Theorem
If f is continuous on [a,b] and N is any number between f(a) and f(b), then f(c) = N for some c in [a,b].
The theorem is a precise version of "a continuous curve cannot jump over a value without passing through it." It fails instantly for a discontinuous function: a jump discontinuity can skip an entire range of N values without ever equalling any of them.
Example 3.3 — Locating a root
Show that f(x) = x3 − x − 1 has a root between x = 1 and x = 2.
Step 1 — Confirm continuity
f is a polynomial, continuous everywhere, hence on [1,2].
Step 2 — Evaluate the endpoints
f(1) = 1 − 1 − 1 = −1; f(2) = 8 − 2 − 1 = 5
Step 3 — Apply the IVT with N = 0
0 lies between −1 and 5, so by the IVT there exists c ∈ (1,2) with f(c) = 0.
f has a root in (1, 2). (The IVT guarantees existence, not the exact value.)
3.7 Infinite Limits and Vertical Asymptotes
When f(x) grows without bound as x → a, we write
— a statement describing behaviour, not a claim that the limit exists as a number. If limx→a− f(x) or limx→a+ f(x) is ±∞, the line x = a is a vertical asymptote of the graph. For a rational function in lowest terms, vertical asymptotes occur exactly at the zeros of the denominator (which are not also zeros of the numerator — those give removable discontinuities instead, §3.5).
Example 3.4 — An infinite limit
Describe the behaviour of f(x) = 1/(x−2)2 as x → 2.
Step 1 — Examine the denominator near x = 2
(x−2)2 → 0+ from both sides (a square is never negative), so 1/(x−2)2 grows without bound from both sides.
Step 2 — State the one-sided and two-sided results
limx→2− f(x) = limx→2+ f(x) = ∞, so limx→2 f(x) = ∞ and x = 2 is a vertical asymptote.
3.8 Limits at Infinity and Horizontal Asymptotes
The mirror question asks what happens to f(x) as x itself grows without bound:
If either holds, the line y = L is a horizontal asymptote. The key building-block limit, provable from the precise definition, is:
which says simply that 1 divided by a larger and larger number shrinks to 0.
3.9 End Behavior of Rational Functions
To evaluate a limit at infinity of a rational function, divide numerator and denominator by the highest power of x appearing in the denominator, then apply Eq. 3.6 term by term.
Example 3.5 — Limit at infinity of a rational function
Evaluate limx→∞ 3x2 − x5x2 + 2.
Step 1 — Divide every term by x², the highest power in the denominator
3 − 1/x5 + 2/x2
Step 2 — Apply Eq. 3.6 to each term with x in it
As x → ∞, 1/x → 0 and 2/x2 → 0.
3 − 05 + 0 = 3/5, so y = 3/5 is a horizontal asymptote.
Comparing the degrees of numerator and denominator predicts the answer without dividing: if the degrees are equal, the limit is the ratio of leading coefficients (as above, 3/5); if the numerator's degree is smaller, the limit is 0; if larger, the function grows without bound and there is no horizontal asymptote (though there may be a slant one, beyond this course's scope).
3.10 Worked Capstone Example
Example 3.6 — Full asymptote and continuity analysis
For f(x) = 2xx − 1, find all vertical and horizontal asymptotes, and state where f is continuous.
Step 1 — Vertical asymptote (§3.7)
The denominator is zero at x = 1, and the numerator is not, so x = 1 is a vertical asymptote. As x → 1+, x−1 → 0+ so f(x) → +∞; as x → 1−, f(x) → −∞.
Step 2 — Horizontal asymptote (§3.8–3.9)
Numerator and denominator both have degree 1, so the limit at ±∞ is the ratio of leading coefficients: 2/1 = 2.
Vertical asymptote: x = 1. Horizontal asymptote: y = 2.
Step 3 — Continuity (§3.3–3.4)
f is a rational function, continuous everywhere on its domain, which is every real number except x = 1.
3.11 Nomenclature
| Symbol | Meaning |
|---|---|
| a+, a− | Approaching a from the right / left |
| IVT | Intermediate Value Theorem |
| ∞, −∞ | Symbols describing unbounded growth, not numbers |
| Vertical asymptote | A line x = a that f approaches infinitely closely as one or both one-sided limits are ±∞ |
| Horizontal asymptote | A line y = L that f(x) approaches as x → ±∞ |
3.12 Problems
- One-sided limits. For f(x) = ⌊x⌋ (the floor/greatest-integer function), find limx→2− f(x) and limx→2+ f(x). Does the two-sided limit exist?
- Continuity check. Let g(x) = x2 for x ≤ 1 and g(x) = 2x for x > 1. Is g continuous at x = 1? Justify using all three conditions of §3.3.
- Classify the discontinuity. For h(x) = 1/(x−4), classify the discontinuity at x = 4.
- Removable discontinuity. Let k(x) = (x2−1)/(x−1) for x ≠ 1. What value should be assigned to k(1) to make k continuous there?
- IVT application. Show that cos x = x has a solution in (0, π/2).
- IVT, why continuity matters. Give an example of a discontinuous function on [0,1] with f(0) = −1 and f(1) = 1 that never equals 0, and explain why this does not contradict the IVT.
- Vertical asymptote. Find all vertical asymptotes of f(x) = (x+1)/(x2−4), and state the sign of f on each side of each asymptote.
- Horizontal asymptote, degree equal. Find limx→∞ (4x3−1)/(2x3+x).
- Horizontal asymptote, numerator smaller degree. Find limx→−∞ (5x+1)/(x2−3).
- No horizontal asymptote. Explain why f(x) = (x2+1)/(x+1) has no horizontal asymptote, using the degree-comparison shortcut of §3.9.
- Combined. For f(x) = (3x−6)/(x2−4), determine whether x = 2 gives a vertical asymptote or a removable discontinuity, and find the horizontal asymptote.
3.13 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §2.4 (one-sided limits), §2.5 (continuity), §2.6 (limits involving infinity; asymptotes of graphs).
- Pearson MyLab Math — the Week 3 homework set is assigned directly from the sections above; see the Calendar for the due date.