Difficulty: Intermediate | Prerequisites: Rates of Change notes, Limit Calculation Techniques notes
This document covers four topics that together round out the foundations of limits in Calculus I. The epsilon-delta definition makes the intuitive idea of a limit rigorous. Continuity tells you when a function behaves "nicely" enough that limits are just plug-and-chug. The Intermediate Value Theorem turns continuity into a tool for proving that equations have solutions. Limits involving infinity give you asymptotes, which describe how functions behave at the extremes of their domain. These ideas recur throughout calculus and analysis.
The formal (ε-δ) definition of a limit says: for every tolerance ε, there is a neighbourhood δ around the point where the function stays within ε of the limit value. Continuous functions let you skip limit calculations entirely. The IVT guarantees that a continuous function hits every value between f(a) and f(b). Asymptotes describe end-behaviour and blow-up points for rational functions.
Epsilon-delta (ε-δ) definition of a limit
lim (x → c) f(x) = L means: for every ε > 0 there exists a δ > 0 such that whenever 0 < |x − c| < δ, we have |f(x) − L| < ε. Think of it as: you name a tolerance (ε), and I find a neighbourhood (δ) around c where the function stays within that tolerance of L.
Continuity at a point
A function f is continuous at x = c if lim (x → c) f(x) = f(c). In simple terms, the function's limit at the point equals its value there, so there is no jump, hole, or blow-up.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b], then f takes on every value between f(a) and f(b) at least once. Think of it as: a continuous curve that starts below a line and ends above it must cross that line somewhere.
Horizontal asymptote
A horizontal line y = L that the graph of f approaches as x → +∞ or x → −∞. In simple terms, the function levels off at L when x gets very large or very negative.
Vertical asymptote
A vertical line x = a where f(x) → ±∞. Think of it as: the function blows up near x = a, shooting off towards positive or negative infinity.
Oblique (slant) asymptote
A non-horizontal, non-vertical line y = mx + b that the graph approaches as x → ±∞, occurring when the numerator's degree is exactly one more than the denominator's.
Removable discontinuity
A point where the limit exists but the function is either undefined or has a different value. In simple terms, a "hole" in the graph that could be patched by redefining f at that one point.
Every ε-δ problem has two stages: scratchwork to find δ in terms of ε, then the formal proof that your δ works.
Worked example (linear function): Prove lim (x → 2) 2x = 4.
We need: 0 < |x − 2| < δ implies |2x − 4| < ε.
Rewrite: |2x − 4| = 2|x − 2|.
So we need 2|x − 2| < ε, i.e. |x − 2| < ε/2.
Choose δ = ε/2.
Proof: let ε > 0 be given, set δ = ε/2. If 0 < |x − 2| < δ, then |2x − 4| = 2|x − 2| < 2δ = ε.
Worked example (reciprocal function): For g(x) = 1/x near x = 1, find δ so that |g(x) − 1| < 0.5.
We need |1/x − 1| < 0.5, so 0.5 < 1/x < 1.5.
Solving: x lies between 2/3 and 2.
The largest symmetric interval about 1 inside (2/3, 2) is (2/3, 4/3).
The smaller distance from 1 is 1/3, so δ = 1/3.
A function is continuous at c when lim (x → c) f(x) = f(c). For continuous functions, evaluating a limit is just substitution.
Functions that are continuous on their domains:
Polynomials (everywhere)
Rational functions (everywhere the denominator is non-zero)
Trigonometric functions (sine and cosine everywhere; tangent where cos x ≠ 0)
Absolute value (everywhere)
Watch for division by zero: a quotient of continuous functions is continuous only where the denominator is non-zero.
If f is continuous on [a, b] and N is any value between f(a) and f(b), then there exists at least one c in (a, b) with f(c) = N.
Worked example: Show that x⁸⁰ + x⁷⁹ = 1 + cos x has a solution in [0, 1].
Define g(x) = 1 + cos x − x⁸⁰ − x⁷⁹.
g(0) = 1 + 1 − 0 − 0 = 2 > 0.
g(1) = 1 + cos 1 − 1 − 1 = cos 1 − 1 < 0 (since cos 1 ≈ 0.54).
g is continuous (it is a sum of continuous functions).
By the IVT, there is some α in (0, 1) where g(α) = 0, which means α solves the original equation.
Horizontal asymptotes come from limits as x → ±∞.
For rational functions p(x)/q(x), compare the degrees of p and q:
deg(p) < deg(q): horizontal asymptote at y = 0.
deg(p) = deg(q): horizontal asymptote at y = (leading coefficient of p) / (leading coefficient of q).
deg(p) = deg(q) + 1: oblique asymptote (find it by polynomial long division).
Worked example: lim (x → ∞) (x⁴ + 3x²) / (x⁴ − 22x + 60).
Divide every term by x⁴: (1 + 3/x²) / (1 − 22/x³ + 60/x⁴) → 1.
Horizontal asymptote at y = 1.
Vertical asymptotes occur where the denominator is zero and the numerator is not (after cancellation).
(x − 1)/(x² − 1) has a vertical asymptote at x = −1 only. At x = 1, the factor cancels, leaving a removable discontinuity.
x/(x² − 1) has vertical asymptotes at both x = 1 and x = −1, because nothing cancels.
Formal definition of a limit:
lim (x → c) f(x) = L ⟺ for every ε > 0, there exists δ > 0 such that 0 < |x − c| < δ implies |f(x) − L| < ε.
Continuity condition:
f is continuous at c ⟺ lim (x → c) f(x) = f(c).
IVT (statement):
If f is continuous on [a, b] and N is between f(a) and f(b), then there exists c ∈ (a, b) with f(c) = N.
Asymptote rules for rational functions p(x)/q(x):
deg(p) < deg(q) → horizontal asymptote y = 0
deg(p) = deg(q) → horizontal asymptote y = leading coeff of p / leading coeff of q
deg(p) = deg(q) + 1 → oblique asymptote (use long division)
Denominator = 0 and numerator ≠ 0 (after cancellation) → vertical asymptote
The IVT is used in numerical methods (bisection method) to find roots of equations that cannot be solved algebraically. Engineers rely on it to guarantee that a temperature, pressure, or stress value is reached somewhere along a continuous process. Asymptotes describe steady-state behaviour in systems: a chemical reaction rate approaching a maximum (horizontal asymptote), or a resonance frequency where a response blows up (vertical asymptote).
Students often think the ε-δ definition requires finding the "best" δ. Any δ that works is sufficient. A smaller δ is always valid too.
A common error is assuming that if the denominator is zero, there is always a vertical asymptote. If the numerator is also zero at that point (and the factor cancels), you get a removable discontinuity instead.
Students sometimes apply the IVT in reverse, claiming that if f(a) and f(b) have the same sign, there is no root between them. The IVT only guarantees a root when the signs differ; it says nothing when they match.
"Continuous everywhere" does not mean "continuous at every real number." A rational function is continuous on its domain, but its domain excludes the points where the denominator is zero.
⚠️ Expect at least one ε-δ proof, likely for a linear function. Know the two-step process: scratchwork, then proof.
⚠️ Be prepared to identify all asymptotes (horizontal, vertical, oblique) for a given rational function.
⚠️ The IVT is often tested as a "show that a solution exists" problem. You need to find an interval where f changes sign and state that f is continuous.
⚠️ Know which standard function families are continuous on their domains: polynomials, rational functions, trig functions, absolute value.
True or false: in an ε-δ proof, you must always find the largest possible δ. False (any δ that works is fine).
Fill in the blank: a function is continuous at c if lim (x → c) f(x) = ______. f(c).
True or false: the IVT can prove that a solution exists, but it cannot tell you the exact value. True.
Fill in the blank: if deg(numerator) = deg(denominator), the horizontal asymptote is y = ______. The ratio of the leading coefficients.
True or false: (x − 1)/(x² − 1) has a vertical asymptote at x = 1. False (it has a removable discontinuity there; the vertical asymptote is at x = −1).
Q: Using the ε-δ definition, prove that lim (x → 3) (5x − 1) = 14.
A: We need |5x − 1 − 14| < ε, i.e. |5x − 15| = 5|x − 3| < ε, so |x − 3| < ε/5. Set δ = ε/5. If 0 < |x − 3| < δ, then |5x − 15| = 5|x − 3| < 5δ = ε.
Q: Use the IVT to show that x³ + x − 1 = 0 has a solution between 0 and 1.
A: Let f(x) = x³ + x − 1. Then f(0) = −1 < 0 and f(1) = 1 > 0. Since f is a polynomial (hence continuous), the IVT guarantees some c ∈ (0, 1) with f(c) = 0.
Q: Find all asymptotes of f(x) = (2x² + 3) / (x² − 4).
A: Horizontal: deg equals deg, so y = 2/1 = 2. Vertical: x² − 4 = 0 gives x = 2 and x = −2. The numerator is non-zero at both, so vertical asymptotes at x = 2 and x = −2.
Q: Is f(x) = (x² − 9)/(x − 3) continuous at x = 3? Explain.
A: f is not defined at x = 3 (0/0), so it is not continuous there. The limit exists (factor to get x + 3, which equals 6 at x = 3), so it is a removable discontinuity.
Q: Does x/(x² − 1) have a removable discontinuity or a vertical asymptote at x = 1?
A: The numerator at x = 1 is 1 (non-zero), while the denominator is 0. No factor cancels. Vertical asymptote at x = 1.
The ε-δ definition reappears when you study uniform continuity and sequences in analysis courses. The IVT is the theoretical backbone of root-finding algorithms such as the bisection method. Asymptotes connect to curve sketching (using first and second derivatives), which is a major topic later in MTH 132.
Epsilon-delta definition, ε-δ proof, formal limit definition, continuity, continuous function, Intermediate Value Theorem, IVT, existence of roots, horizontal asymptote, vertical asymptote, oblique asymptote, slant asymptote, removable discontinuity, hole in graph, rational function asymptotes, degree comparison, MTH 132, Calculus I, Michigan State University, Sections 2.4-2.6