Maths I — Week 3: Limits and Continuity

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Maths I — Week 3: Limits and Continuity

The Big Idea: A limit describes where a function is headed, not necessarily where it is. This distinction between "approaching" and "arriving" is the core conceptual leap into calculus. Continuity is simply the property that the function actually arrives where it's headed.

1. The Intuitive Limit

lim⁡x→af(x)=L\lim_{x \to a} f(x) = L
Means: As xx gets arbitrarily close to aa (from either side), f(x)f(x) gets arbitrarily close to LL.
The limit only cares about the
approach to aa, not what happens at aa itself. A function can have a limit at a point even if it's undefined there.

1.1 One-Sided Limits

NotationMeaning
lim⁡x→a+f(x)\lim_{x \to a^+} f(x)Right-Hand Limit (approaching from a+ϵa + \epsilon)
lim⁡x→a−f(x)\lim_{x \to a^-} f(x)Left-Hand Limit (approaching from a−ϵa - \epsilon)
The two-sided limit exists if and only if LHL = RHL.

2. The Formal (ϵ,δ)(\epsilon, \delta) Definition

For every epsilon>0\\epsilon > 0, there exists delta>0\\delta > 0 such that: 0<∣x−a∣<δ  ⟹  ∣f(x)−L∣<ϵ0 < |x - a| < \delta \implies |f(x) - L| < \epsilon
Plain English: "Give me any target tolerance ϵ\epsilon around the output, and I'll find a corresponding input window δ\delta around aa that guarantees the output stays within that tolerance."

3. Limit Laws and Algebra

If lim⁡x→af(x)=L\lim_{x\to a}f(x) = L and lim⁡x→ag(x)=M\lim_{x\to a}g(x) = M:
LawFormula
Sumlim⁡[f+g]=L+M\lim [f+g] = L + M
Productlim⁡[f⋅g]=L⋅M\lim [f \cdot g] = L \cdot M
Quotientlim⁡[f/g]=L/M\lim [f/g] = L/M, provided M≠0M \neq 0
Powerlim⁡[fn]=Ln\lim [f^n] = L^n

4. Indeterminate Forms and Factoring

When direct substitution gives 00\frac{0}{0}, factor and cancel.

Abstract Solution (Strategy) — 0/00/0 forms

  1. [Factor]: Factor numerator and denominator.
  2. [Cancel]: Cancel the common factor (x−a)(x - a).
  3. [Substitute]: Now the plug-in works.
Worked Example:
lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}
  • Direct sub: 0/00/0.
  • Factor: (x−2)(x+2)(x−2)=x+2\frac{(x-2)(x+2)}{(x-2)} = x+2.
  • Substitute: 2+2=42 + 2 = 4.
  • Result: 4.

5. Standard Limits (Memorize These)

LimitValue
lim⁡x→0sin⁡xx\lim_{x \to 0} \frac{\sin x}{x}11
lim⁡x→01−cos⁡xx\lim_{x \to 0} \frac{1 - \cos x}{x}00
lim⁡x→0ex−1x\lim_{x \to 0} \frac{e^x - 1}{x}11
lim⁡x→0ln⁡(1+x)x\lim_{x \to 0} \frac{\ln(1+x)}{x}11
lim⁡x→∞(1+1n)n\lim_{x \to \infty} \left(1 + \frac{1}{n}\right)^nee
lim⁡x→∞nn!n\lim_{x \to \infty} \frac{n}{\sqrt[n]{n!}}ee
The standard limit
lim⁡u→0eu−1u=1\lim_{u \to 0} \frac{e^u - 1}{u} = 1 is extremely powerful. Any limit of the form n(e1/f(n)−1)n(e^{1/f(n)} - 1) as n→∞n \to \infty can be resolved by substituting u=1/f(n)→0u = 1/f(n) \to 0 and multiplying/dividing to match the form.

6. L'Hôpital's Rule

Used for: Indeterminate forms 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} that can't be factored.
lim⁡x→cf(x)g(x)=lim⁡x→cf′(x)g′(x)\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
Critical Note: You differentiate the numerator and denominator separately — do NOT use the Quotient Rule.
L'Hôpital only applies when the original limit gives
0/00/0 or ∞/∞\infty/\infty. Don't apply it to limits that already have a finite value after direct substitution.

7. Continuity — The Three Conditions

A function ff is continuous at x=ax = a if and only if all three hold:
  1. f(a)f(a) is defined.
  2. lim⁡x→af(x)\lim_{x \to a} f(x) exists (LHL = RHL).
  3. lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a) — the limit equals the function value.
All three conditions must hold simultaneously. A function failing even one is discontinuous at
aa.

7.1 Types of Discontinuity

TypeWhat failsExample
Removablef(a)f(a) ≠ limit, or f(a)f(a) undefinedx2−1x−1\frac{x^2-1}{x-1} at x=1x=1 (a "hole")
JumpLHL ≠ RHL (both finite)Floor function ⌊x⌋\lfloor x \rfloor at integers
InfiniteLimit = ±∞\pm\infty1x\frac{1}{x} at x=0x=0 (vertical asymptote)

8. Pattern — Piecewise Continuity Check

What to recognize: A piecewise function asking for the value of a constant kk that makes it continuous.

Abstract Solution (Strategy)

  1. [Identify the break point]: Usually the xx-value where the formula changes.
  2. [Compute LHL and RHL]: Evaluate from each piece.
  3. [Equate to f(a)]: Set LHL = RHL = f(a)f(a) and solve for kk.
Worked Example:
Find m+nm+n if f(x)={3mx+nx<111x=15mx+2nx>1f(x) = \begin{cases} 3mx+n & x < 1 \\ 11 & x = 1 \\ 5mx+2n & x > 1 \end{cases} is continuous at x=1x=1.
  • LHL: 3m+n=113m + n = 11.
  • RHL: 5m+2n=115m + 2n = 11.
  • Solve: From (1), n=11−3mn = 11 - 3m. Sub into (2): 5m+22−6m=11  ⟹  m=115m + 22 - 6m = 11 \implies m = 11.
  • n=11−33=−22n = 11 - 33 = -22.
  • m+n=11−22=−11m + n = 11 - 22 = -11.
  • Result: −11-11.

9. The Intermediate Value Theorem (IVT)

Statement: If ff is continuous on [a,b][a, b] and kk is any value between f(a)f(a) and f(b)f(b), then there exists at least one c∈(a,b)c \in (a, b) such that f(c)=kf(c) = k.
Power Move: To prove a root exists in [a,b][a, b]:
  1. Show ff is continuous on [a,b][a, b].
  2. Show f(a)<0f(a) < 0 and f(b)>0f(b) > 0 (or vice versa).
  3. By IVT, f(c)=0f(c) = 0 for some cc in between.

10. Floor Function Limits (⌊x⌋\lfloor x \rfloor)

DirectionExampleValue
Right-hand (a+a^+)lim⁡x→4+⌊x⌋\lim_{x \to 4^+} \lfloor x \rfloor44 (just above 4)
Left-hand (a−a^-)lim⁡x→4−⌊x⌋\lim_{x \to 4^-} \lfloor x \rfloor33 (just below 4 is still 3)
Key rule: At any integer nn, the floor function has a jump discontinuity: LHL = n−1n-1, RHL = nn.

11. Common Mistakes

MistakeWhy it happensCorrect approach
Applying L'Hôpital to non-indeterminate formsHabit of always differentiating.Check: does direct substitution give 0/00/0 or ∞/∞\infty/\infty? If not, don't use L'Hôpital.
Treating ⌊a−rfloor=a\lfloor a^- \\rfloor = aForgetting that "just below an integer" rounds down.lim⁡x→n−⌊xrfloor=n−1\lim_{x \to n^-} \lfloor x \\rfloor = n - 1 always.
Using the Quotient Rule in L'HôpitalThe idea is to differentiate numerator and denominator separately.Compute f′(x)f'(x) and g′(x)g'(x) individually, then form the new fraction.
Confirming continuity without all 3 checksChecking only that the limit exists.All three: defined, limit exists, limit = value.

12. Flashcards

<Flashcard front="What are the three conditions for continuity at x = a?" back="1. f(a) is defined. 2. lim f(x) as x→a exists. 3. The limit equals f(a)." /> <Flashcard front="What is the limit of (sin x)/x as x→0?" back="1. This is a standard limit — memorize it." /> <Flashcard front="What is a removable discontinuity?" back="The limit exists but f(a) is either undefined or doesn't equal the limit. Appears as a 'hole' in the graph." /> <Flashcard front="State the Intermediate Value Theorem." back="If f is continuous on [a,b], then for any value k between f(a) and f(b), there exists c in (a,b) with f(c) = k." /> <Flashcard front="Why does lim[x→4⁻] floor(x) = 3, not 4?" back="Approaching 4 from the left means x is like 3.999..., and the floor of 3.999 is 3, not 4." />

13. Practice Targets

  • Evaluate lim⁡x→0e3x−1x\lim_{x \to 0} \frac{e^{3x} - 1}{x} using the standard limit identity.
  • Determine continuity at every integer for g(x)=⌊2x⌋g(x) = \lfloor 2x \rfloor.
  • Verify a root exists for h(x)=x3−x−1h(x) = x^3 - x - 1 in [1,2][1, 2] using IVT.
  • Attempt Week 3 Questions 11–13 (all limit/continuity based).

14. Connections

Connects toHow
Week 4 — DerivativesThe derivative is defined as a limit. Mastering limits is prerequisite to differentiation.
Week 7 — SequencesSequence limits use the same degree-ratio and algebraic combination techniques.
Week 8 — L'HôpitalQuestions 8 explicitly applies L'Hôpital three times to a sin/polynomial expression.

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