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Functions and limits: Class 12 Mathematics
Functions, domain and range, composite and inverse functions, the idea of a limit, limit theorems, important limits and continuity, with worked examples.
Key points
A function f from a set A to a set B is a rule that assigns to each element x of A exactly one element y of B. We write y = f(x). A is the domain, and the set of values that f takes is the range. A vertical line meets the graph of a function at only one point.
Types of function include polynomial (such as x² − 3x + 2), rational (a ratio of polynomials), exponential (aˣ), logarithmic (logₐ x) and trigonometric functions. The domain of a rational function excludes the values that make the denominator zero, and the domain of a square root function requires the quantity under the root to be zero or positive.
A function is even if f(−x) = f(x), and its graph is symmetric about the y axis, for example x² and cos x. It is odd if f(−x) = −f(x), and its graph is symmetric about the origin, for example x³ and sin x.
The composite function (f ∘ g)(x) = f(g(x)) is found by putting g(x) in place of x in f. In general f(g(x)) is not the same as g(f(x)).
A function is one to one if different inputs give different outputs. Such a function has an inverse f⁻¹, found by writing y = f(x), solving for x in terms of y and then interchanging x and y. The graph of f⁻¹ is the mirror image of the graph of f in the line y = x, and f(f⁻¹(x)) = x.
The limit of f(x) as x approaches a is L if f(x) can be made as close to L as we like by taking x close enough to a, but not equal to a. We write lim f(x) = L as x → a. The limit exists only if the left hand and right hand limits are equal.
Limit theorems. If lim f(x) = L and lim g(x) = M as x → a, then:
- the limit of f(x) + g(x) is L + M, and of f(x) − g(x) is L − M
- the limit of f(x)g(x) is LM
- the limit of f(x)/g(x) is L/M, provided that M is not zero
- the limit of [f(x)]ⁿ is Lⁿ, and the limit of k f(x) is kL for a constant k
To find a limit that gives 0/0 on direct substitution, factorise and cancel, or rationalise, and then substitute. For example, the limit of (x² − 9)/(x − 3) as x → 3 equals the limit of (x + 3), which is 6.
Important limits:
- lim (sin x)/x = 1 as x → 0, when x is in radians
- lim (1 − cos x)/x = 0 as x → 0
- lim (1 + 1/n)ⁿ = e as n → ∞, where e ≈ 2.718
- lim (1 + x)^(1/x) = e as x → 0
- lim (aˣ − 1)/x = ln a as x → 0, and lim (eˣ − 1)/x = 1
A function f is continuous at x = a if f(a) is defined, the limit of f(x) as x → a exists, and the limit equals f(a). A polynomial is continuous everywhere. A function is discontinuous where it is not defined, where the limit does not exist, or where the limit differs from the value.
Study tip: For every limit, first try direct substitution. If you get 0/0, decide whether to factorise, rationalise or use a standard limit, and show each step.
Definitions and flashcards
Say the meaning first, then open the card to check.
Function
A rule that assigns to each element of the domain exactly one element of the range.
Domain
The set of all allowed inputs of a function.
Range
The set of all values that the function takes.
Composite function
The function f(g(x)), formed by applying g first and then f.
Inverse function
The function that reverses f, so that f⁻¹(f(x)) = x.
Limit
The value that f(x) approaches as x approaches a given number.
Continuity
A function is continuous at a if its limit at a exists and equals its value there.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Composite function | (f ∘ g)(x) = f(g(x)) | Apply g first, then f. |
| Limit of sin x / x | lim (sin x / x) = 1 | x → 0, x in radians. |
| Number e | lim (1 + 1/n)ⁿ = e | n → ∞. |
| Exponential limit | lim (eˣ − 1) / x = 1 | x → 0. |
| Difference of squares | (x² − a²) / (x − a) = x + a | Used to remove 0/0 forms. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
Find the domain of f(x) = √(x − 2) / (x − 5). (2 marks)
The quantity under the root must be non-negative, so x ≥ 2. The denominator must not be zero, so x ≠ 5. The domain is x ≥ 2 with x ≠ 5.
Show that f(x) = x³ + x is an odd function. (2 marks)
f(−x) = (−x)³ + (−x) = −x³ − x = −(x³ + x) = −f(x), so f is odd.
Find f ∘ g and g ∘ f if f(x) = 2x + 1 and g(x) = x². (3 marks)
f(g(x)) = 2x² + 1. g(f(x)) = (2x + 1)² = 4x² + 4x + 1. They are different.
Find the inverse of f(x) = 3x − 4. (2 marks)
Put y = 3x − 4, so x = (y + 4)/3. Interchanging x and y gives f⁻¹(x) = (x + 4)/3.
State the conditions for continuity of f at x = a. (3 marks)
f(a) must exist, the limit of f(x) as x → a must exist, and the limit must equal f(a).
Numericals with worked steps
Evaluate the limit of (x² − 16)/(x − 4) as x → 4.
- Direct substitution gives 0/0.
- Factorise: x² − 16 = (x − 4)(x + 4).
- Cancel the factor x − 4: the expression is x + 4.
- Put x = 4: the limit is 8.
Answer: 8
Evaluate the limit of sin 3x / x as x → 0.
- Multiply and divide by 3: (sin 3x / 3x) × 3.
- As x → 0, 3x → 0, and sin 3x / 3x → 1.
- The limit is 1 × 3 = 3.
Answer: 3
Evaluate the limit of (√(x + 4) − 2)/x as x → 0.
- Direct substitution gives 0/0. Rationalise by multiplying top and bottom by √(x + 4) + 2.
- The numerator becomes (x + 4) − 4 = x.
- The expression is x / [x(√(x + 4) + 2)] = 1 / (√(x + 4) + 2).
- At x = 0 this is 1/(2 + 2) = 1/4.
Answer: 1/4
Evaluate the limit of (1 + 2/x)^x as x → ∞.
- Put x = 2n, so 2/x = 1/n. As x → ∞, n → ∞.
- (1 + 1/n)^(2n) = [(1 + 1/n)ⁿ]².
- The limit is e².
Answer: e²
Common mistakes
- Substituting into 0/0 and stating the limit as 0 or undefined. A 0/0 form needs factorising or rationalising.
- Using sin x / x → 1 with the angle in degrees. It requires radians.
- Writing that f(g(x)) equals g(f(x)).
- Forgetting to exclude the values that make a denominator zero when finding a domain.
- Stating that a limit exists when the left and right limits are different.