Difficulty: Introductory to Intermediate | Prerequisites: Limit definitions and evaluation (Sections 1.1–1.2)
Once you know the basic definition of a limit and how to evaluate simple ones, you need techniques for trickier functions: radicals that give 0/0, piecewise functions with different rules on each side, and the standard trig functions whose domains and behaviours you will need throughout calculus. This section also covers end behaviour (what happens as x goes to ±∞) for odd and even functions, and introduces the key limits of exponential, logarithmic, and trigonometric functions that you should know by sight.
When a radical gives 0/0, multiply by the conjugate to clear it. Piecewise limits require checking the correct piece from each side. End behaviour for odd functions differs depending on direction, while even functions behave the same in both directions. Memorise the domains and key limit behaviours of the six trig functions.
Conjugate
The conjugate of a two-term expression is the same expression with the sign between the terms flipped. For example, the conjugate of (a − √b) is (a + √b). In simple terms, you flip the sign in front of the radical to create a difference-of-squares situation that eliminates the square root.
Piecewise function
A function defined by different rules on different intervals. When evaluating limits of a piecewise function, you must use the correct piece for each side you are approaching from. Think of it as: the left-hand limit uses the piece that governs values just below a, and the right-hand limit uses the piece just above a.
End behaviour
What a function does as x→+∞ or x→−∞. For polynomials, end behaviour is entirely determined by the leading term. In simple terms, zoom out far enough on the graph and only the highest-power term matters.
Odd function
A function satisfying f(−x) = −f(x). Its graph has rotational symmetry about the origin. End behaviour: the two ends go in opposite directions (one to +∞, the other to −∞).
Even function
A function satisfying f(−x) = f(x). Its graph is symmetric about the y-axis. End behaviour: both ends go in the same direction.
Odd functions (positive leading coefficient):
As x→+∞, f(x)→+∞
As x→−∞, f(x)→−∞
Odd functions (negative leading coefficient): the pattern reverses.
Even functions (positive leading coefficient):
As x→+∞, f(x)→+∞
As x→−∞, f(x)→+∞
Even functions (negative leading coefficient): both ends go to −∞.
As x→+∞ or x→−∞, limits of √(ax + c) behave like √(ax) for large x because the constant becomes negligible.
To check signs when x→−∞, note that √(x²) = |x|. When x is negative, |x| = −x.
Example: lim (x→+∞) √(x² + 2) = |x| → x (positive), so the function grows without bound.
Example: lim (x→−∞) √(x² + 2) = |x| → −x (because x is negative), so √(x²+2) still grows, but you need to track the sign carefully in context.
When direct substitution of a limit involving a radical gives 0/0, multiply the numerator and denominator by the conjugate of the radical expression.
This creates a difference of squares, eliminating the radical from either the numerator or denominator.
Then cancel the common factor and substitute.
Example: lim (x→4) (√x − 2) / (x − 4)
Direct substitution: 0/0.
Multiply by (√x + 2) / (√x + 2).
Numerator becomes x − 4 (difference of squares).
Cancel (x − 4): left with 1 / (√x + 2).
Plug in x = 4: 1 / (2 + 2) = 1/4.
When approaching from the left (x→a⁻), use the piece of the function that applies for x < a.
When approaching from the right (x→a⁺), use the piece that applies for x > a.
If the two pieces give different values at x = a, the two-sided limit does not exist.
Do not just look at x = a in the piecewise definition. Look at which piece governs values near a from each side.
Limits in odd functions: the function's behaviour on one side of the origin mirrors (with sign flip) its behaviour on the other side.
Limits in even functions: the function's behaviour is identical on both sides of the origin.
sin(x): period 2π, domain all real numbers
cos(x): period 2π, domain all real numbers
tan(x): period π, domain excludes x = π/2 + nπ (where n is any integer)
cot(x): period π, domain excludes x = nπ
sec(x): period 2π, domain excludes x = π/2 + nπ
csc(x): period 2π, domain excludes x = nπ
lim (x→0⁺) ln(x) = −∞
lim (x→+∞) ln(x) = +∞
lim (x→−∞) eˣ = 0
lim (x→+∞) eˣ = +∞
lim (x→0⁺) 1/x = +∞
lim (x→0⁻) 1/x = −∞
lim (x→0) sin(x)/x = 1 (not from this chapter, but frequently useful)
The compound interest limit: lim (n→∞) (1 + r/n)^n = eʳ. This is the continuous compounding formula.
Graphs of ln(x), eˣ, and 1/x have characteristic shapes and asymptotic behaviours you should be able to sketch from memory.
Conjugate identity (difference of squares):
(a - \sqrt{b})(a + \sqrt{b}) = a^2 - bCompound interest limit:
\lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^n = e^rTrigonometric function domains (values excluded):
Function | Period | Domain excludes |
|---|---|---|
sin x | 2π | (nothing excluded) |
cos x | 2π | (nothing excluded) |
tan x | π | x = π/2 + nπ |
cot x | π | x = nπ |
sec x | 2π | x = π/2 + nπ |
csc x | 2π | x = nπ |
Students often multiply by the conjugate but forget to multiply both the numerator and the denominator. You must multiply by conjugate/conjugate (which equals 1) so the expression's value does not change.
Students sometimes use the wrong piece of a piecewise function when evaluating a one-sided limit. The left-hand limit uses the piece for x < a, and the right-hand limit uses the piece for x > a, regardless of what happens at x = a itself.
Students often think lim (x→∞) sin(x) exists. It does not. Sine oscillates between −1 and 1 forever and never settles on a single value.
Students sometimes confuse √(x²) with x. In fact, √(x²) = |x|, which equals x when x ≥ 0 and −x when x < 0. This matters for limits as x→−∞.
⚠️ Conjugate problems appear regularly. Know how to spot them: a radical in the numerator or denominator, and direct substitution giving 0/0.
⚠️ Piecewise limit questions are popular because they test whether you understand one-sided limits mechanically, not just conceptually.
⚠️ Trig function domains are tested directly. If you are asked "where is tan(x) undefined?" you need the answer immediately: x = π/2 + nπ.
⚠️ Know the shapes of eˣ, ln(x), and 1/x cold. End behaviour questions on these functions are quick marks if you can sketch them from memory.
Fill in the blank: The conjugate of (3 − √7) is __________.
True or False: For an even function with a positive leading coefficient, as x→−∞, f(x)→−∞.
Fill in the blank: tan(x) is undefined at x = __________.
True or False: When evaluating lim (x→2⁻) for a piecewise function, you use the piece defined for x > 2.
Fill in the blank: lim (x→0⁺) ln(x) = __________.
Answers: 1. (3 + √7). 2. False (both ends go to +∞ for a positive leading coefficient). 3. π/2 + nπ (where n is any integer). 4. False (use the piece for x < 2). 5. −∞.
Q: Evaluate lim (x→9) (√x − 3) / (x − 9) using the conjugate technique.
A: Direct substitution gives 0/0. Multiply numerator and denominator by (√x + 3). The numerator becomes x − 9 (difference of squares). Cancel (x − 9) to get 1/(√x + 3). Plug in x = 9: 1/(3 + 3) = 1/6.
Q: A piecewise function is defined as f(x) = 2x + 1 for x < 3, and f(x) = x² for x ≥ 3. Find lim (x→3) f(x).
A: Left-hand limit (x→3⁻): use 2x + 1 → 2(3) + 1 = 7. Right-hand limit (x→3⁺): use x² → 9. Since 7 ≠ 9, the two-sided limit does not exist.
Q: What is the end behaviour of f(x) = −x⁵ + 3x² − 1?
A: The leading term is −x⁵. This is an odd-degree polynomial with a negative leading coefficient. As x→+∞, f(x)→−∞. As x→−∞, f(x)→+∞.
Q: Where is csc(x) undefined?
A: csc(x) = 1/sin(x), so it is undefined wherever sin(x) = 0. That occurs at x = nπ, where n is any integer.
Q: What is lim (x→−∞) eˣ?
A: 0. The exponential function eˣ approaches 0 as x goes to negative infinity but never reaches it.
The conjugate technique reappears when you learn the formal definition of the derivative, where difference quotients involving radicals produce 0/0 forms.
Piecewise function limits are directly tied to continuity (Section 1.5): a piecewise function is continuous at a boundary point only if the limit from both sides equals the function value defined there.
Trig function domains become critical in integration (later in the course), where you need to know where integrands blow up.
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