§ 01The Intuition — Approaching But Not Touching
Calculus is the mathematics of change and accumulation. But before we can differentiate or integrate anything, we need a single foundational idea: the limit.
Imagine walking towards a door. You get halfway there, then halfway of that distance, then half again: always halving the remaining gap. In theory, you never actually reach the door, but it is perfectly clear which point you are heading towards. A limit captures exactly that idea: the value a function tends toward, regardless of whether it ever actually gets there.
Consider the function f(x) = x² + 1. As x gets closer and closer to 2, what happens to f(x)? Let us build a table of values, approaching x = 2 from both sides:
| x | f(x) = x² + 1 |
|---|---|
| 1.0 | 2.000 |
| 1.5 | 3.250 |
| 1.8 | 4.240 |
| 1.9 | 4.610 |
| 1.99 | 4.960 |
| 1.999 | 4.996 |
| → 2 ← | → 5 ← |
| 2.001 | 5.004 |
| 2.01 | 5.040 |
| 2.1 | 5.410 |
| 2.5 | 7.250 |
f(x) approaches 5 from both sides as x → 2
From both the left and the right, f(x) converges to 5. We say: "the limit of f(x) as x approaches 2 equals 5." For this particular function we could have just plugged in x = 2 directly: but as we will soon see, limits become indispensable precisely when direct substitution fails.
§ 02Formal Limit Notation
The standard notation for a limit is:
Reading this aloud: "lim, x approaches a, of f of x, equals L." Each symbol has a precise role:
lim
The limit operator. Instructs us to analyse the behaviour of the expression that follows as the subscript condition is met.
limx→aRead as "the limit as x approaches a of…"
x → a
The approach condition. x is a variable that gets arbitrarily close to a but is never equal to a. The arrow → means "tends toward" or "approaches."
x → 2 means x = 1.9, 1.99, 1.999, …x gets infinitely close but never equals 2.
f(x)
The function being examined. This is the expression whose output behaviour we are tracking as the input x moves toward a.
f(x) = x² + 1, sin(x), (x²−9)/(x−3)…Any function: polynomial, rational, trig, piecewise.
= L
The limit value. L is the unique number that f(x) approaches. If no such number exists, the limit does not exist (DNE).
L ∈ ℝ, or L = ±∞, or DNEInfinite limits and non-existent limits are both valid outcomes.
§ 03Limits That Exist vs. Limits That Don't
Not every limit exists. There are three common scenarios where a limit fails to exist at a point x = a:
When Does a Limit Fail to Exist?
- Jump discontinuity: The left-hand and right-hand limits both exist but are not equal (e.g. a step function). The function "jumps" at x = a.
- Infinite / vertical asymptote: f(x) grows without bound as x → a. We say the limit is ±∞. Technically the limit does not exist as a real number, though we write lim f(x) = ∞ informally.
- Oscillation: The function oscillates infinitely fast without settling on any value (e.g. sin(1/x) as x → 0). Neither left nor right limit exists.
Removable Discontinuity (Hole)
The limit exists, but f(a) either doesn't exist or has the wrong value. The graph has a "hole" at (a, L). Classic example: (x²−9)/(x−3) at x = 3.
Limit exists. f(a) is wrong or missing.
Jump Discontinuity
The left-hand and right-hand limits exist but disagree. The graph "jumps" at x = a. Common in piecewise functions and the floor function.
Left ≠ Right ⟹ Limit DNE.
§ 04Direct Substitution Property
For many common functions, evaluating a limit is as simple as plugging in x = a. This works whenever f is continuous at a, which includes all polynomials, rational functions (where the denominator is non-zero), trig functions, exponentials, and logarithms within their domains.
Simply replace x with a and evaluate. No algebra required.
The function x² + 2x − 1 is a polynomial, continuous everywhere.
sin(x) is continuous everywhere.
Denominator at x = 2: 2 + 1 = 3 ≠ 0, so direct substitution is valid.
§ 05Limits with Algebra — Removing the 0/0 Form
When direct substitution produces 0/0, the fraction is not meaningless, it signals a common factor that can be cancelled. The three main techniques are factoring, rationalising, and simplifying complex fractions.
Technique 1 — Factoring
Factor the numerator (or denominator) so the problematic factor cancels. The cancelled version is equal to the original for all x ≠ a, which is exactly the domain relevant to the limit.
Direct substitution: (9 − 9)/(3 − 3) = 0/0. Indeterminate: factor instead.
- Factor numerator: x² − 9 = (x + 3)(x − 3)
- Cancel (x − 3) for x ≠ 3: (x + 3)(x − 3)/(x − 3) = x + 3
- Now substitute: limx→3 (x + 3) = 3 + 3 = 6
Direct substitution gives 0/0. Factor the numerator.
- x² + 5x + 6 = (x + 2)(x + 3)
- Cancel (x + 2): result is x + 3
- limx→−2 (x + 3) = −2 + 3 = 1
Technique 2 — Rationalising (Conjugates)
When a square root appears in the numerator or denominator, multiply top and bottom by the conjugate to eliminate the root and reveal a cancellable factor.
Direct substitution: (2 − 2)/(4 − 4) = 0/0. Multiply by conjugate (√x + 2)/(√x + 2).
- Numerator: (√x − 2)(√x + 2) = x − 4
- Expression becomes: (x − 4) / [(x − 4)(√x + 2)]
- Cancel (x − 4) for x ≠ 4: 1/(√x + 2)
- Substitute x = 4: 1/(2 + 2) = 1/4
0/0 form. Multiply by conjugate (√(x+9) + 3)/(√(x+9) + 3).
- Numerator: (x + 9) − 9 = x
- Expression: x / [x(√(x+9) + 3)]
- Cancel x: 1/(√(x+9) + 3)
- Substitute x = 0: 1/(3 + 3) = 1/6
§ 06Limit Laws — The Arithmetic of Limits
Limits obey clean arithmetic rules. If limx→a f(x) = L and limx→a g(x) = M, then:
| Law | Statement | Plain English |
|---|---|---|
| Sum | lim [f + g] = L + M | Limit of a sum = sum of limits |
| Difference | lim [f − g] = L − M | Limit of a difference = difference of limits |
| Constant Multiple | lim [c·f] = c·L | Constants factor out of limits |
| Product | lim [f · g] = L · M | Limit of a product = product of limits |
| Quotient | lim [f/g] = L/M, if M ≠ 0 | Limit of a quotient = quotient of limits (M ≠ 0) |
| Power | lim [f]ⁿ = Lⁿ | Limit of a power = power of the limit |
| Root | lim √f = √L (L ≥ 0) | Limit of a root = root of the limit |
Apply laws step by step using limx→2 x = 2:
- lim [3x²] = 3·(lim x)² = 3·4 = 12
- lim [5x] = 5·2 = 10
- lim [1] = 1
- Combine: 12 − 10 + 1 = 3
§ 07Further Worked Examples
0/0 at x = 1. Factor using difference of cubes: a³ − b³ = (a−b)(a²+ab+b²).
- x³ − 1 = (x − 1)(x² + x + 1)
- Cancel (x − 1): result is x² + x + 1
- Substitute: 1 + 1 + 1 = 3
This is one of the most important limits in calculus. Direct substitution gives 0/0, and factoring doesn't apply here. The result is established geometrically via the Squeeze Theorem (covered in its own section), but the answer is:
Divide numerator and denominator by the highest power x²:
- Numerator ÷ x²: 3 + 2/x
- Denominator ÷ x²: 5 − 1/x
- As x → ∞: 2/x → 0 and 1/x → 0
- Result: (3 + 0)/(5 − 0) = 3/5
This looks like ∞ − ∞, an indeterminate form. Combine over a common denominator first.
- 1/x − 1/x² = x/x² − 1/x² = (x − 1)/x²
- As x → 0⁺: numerator → −1, denominator → 0⁺. So the expression → −∞.
- As x → 0⁻: numerator → −1, denominator → 0⁺ (x² is always positive). Expression → −∞.
§ 08Common Mistakes
| Mistake | Example of the Error | Correct Approach |
|---|---|---|
| Confusing f(a) with lim f(x) | "limx→2 f(x) = f(2) always" | They are equal only when f is continuous at a. In general, check whether a hole, jump, or asymptote exists. |
| Concluding 0/0 = 0 | lim (x−3)/(x−3) = 0 because "0 over anything is 0" | 0/0 is indeterminate — it could equal any number. Always factor or use another technique. |
| Forgetting to check both sides | Evaluating only from the right and declaring the limit exists | A two-sided limit exists only if left-hand and right-hand limits are equal. |
| Applying the Quotient Law when M = 0 | lim f/g = L/M when M = 0 | If lim g = 0, the Quotient Law does not apply. The limit may be ∞, −∞, or DNE. |
| Cancelling factors without checking x ≠ a | Cancelling (x − a) when x could equal a | Cancellation is valid because the limit definition excludes x = a (we use 0 < |x − a|). |
§ 09Quick Reference — Key Limits to Memorise
| Limit | Value | Technique / Note |
|---|---|---|
| limx→0 sin(x)/x | 1 | Squeeze Theorem — must have x in radians |
| limx→0 (1 − cos x)/x | 0 | Derived from sin(x)/x identity |
| limx→0 (eˣ − 1)/x | 1 | Definition of e; derivative of eˣ at 0 |
| limx→∞ (1 + 1/x)ˣ | e | Defining limit of Euler's number e ≈ 2.718 |
| limx→0⁺ x·ln(x) | 0 | L'Hôpital or rewrite as 0·∞ |
| limx→∞ xⁿ/eˣ | 0 | Exponential dominates any polynomial |
| limx→∞ ln(x)/x | 0 | Power dominates logarithm |
§ 10Practice Quiz — Introduction to Limits
10 questions spanning intuition, notation, direct substitution, and algebraic manipulation. Immediate feedback on every answer.
Find limx→3 (x² − 9)/(x − 3)
Factor: (x²−9)/(x−3) = (x+3)(x−3)/(x−3) = x+3. Substitute x=3: 3+3 = 6.
What does the notation limx→a f(x) = L mean?
A limit describes the approach behaviour, not the actual value at a. f(a) could equal L, differ from L, or not even be defined.
Evaluate: limx→2 (3x² − 4)
Polynomial is continuous everywhere. Substitute x=2: 3(4) − 4 = 12 − 4 = 8.
Why does direct substitution fail for limx→1 (x² − 1)/(x − 1)?
Plugging x=1 gives (1−1)/(1−1) = 0/0. This is indeterminate, we must factor: (x+1)(x−1)/(x−1) = x+1 → 2.
Find: limx→−3 (x² + 5x + 6)/(x + 3)
Factor numerator: (x+3)(x+2)/(x+3) = x+2. Substitute x=−3: −3+2 = −1.
When does a limit not exist at x = a?
A limit fails to exist when: LHL ≠ RHL (jump), f oscillates without settling, or |f(x)| → ∞. A function being undefined at a does NOT by itself mean the limit fails.
Evaluate using rationalisation: limx→9 (√x − 3)/(x − 9). Enter your answer as a fraction: it equals 1/___.
- Multiply by (√x+3)/(√x+3): numerator = x−9, denominator = (x−9)(√x+3)
- Cancel (x−9): 1/(√x+3)
- Substitute x=9: 1/(3+3) = 1/6
By the Quotient Limit Law, lim [f/g] = (lim f)/(lim g) provided:
The Quotient Law requires lim g ≠ 0. If lim g = 0, division is invalid and the limit must be found by other means.
Evaluate: limx→∞ (5x³ − 2x) / (2x³ + x²)
Divide by x³: (5 − 2/x²)/(2 + 1/x). As x→∞, all 1/xⁿ → 0. Result: 5/2 = 2.5.
The value of limx→0 sin(x)/x (x in radians) is:
This is one of the most important limits in calculus. Proved using the Squeeze Theorem: limx→0 sin(x)/x = 1. It only holds with x in radians.
Quiz complete!