MATH1061 · Mathematics 1a
Limits and Continuity
A limit describes what a function approaches as its input approaches a point — near the point, never necessarily at it. That distinction is the whole game: a limit can exist exactly where the function is undefined, which is why 0/0 forms (the staple exam stem) are answerable by factoring and cancelling rather than by substitution. This is the foundation of the Calculus stream and the first thing the Week-8 Quiz A samples. You must apply the limit laws, recognise the indeterminate forms, use the squeeze (sandwich) theorem to kill oscillating limits, handle one-sided limits, and know what continuity buys you — the right to substitute, plus the Intermediate Value Theorem for root-existence arguments. Everything here is computed by hand on exact values, because that is how it is tested.
What this chapter covers
- 011.1 Three tests you reuse all year (the substitution / factor / squeeze trichotomy)
- 021.2 The function families you must know cold
- 031.3 The limit laws — building big limits from small ones
- 04One-sided limits and when the two-sided limit exists
- 051.4 Reading the picture — limits from a graph
- 06The squeeze (sandwich) theorem
- 071.5 The standard limits worth memorising
- 08Continuity and 1.6 the Intermediate Value Theorem (IVT)
Worked example: a 0/0 limit and a squeeze
- +1(a) Test substitution: at x = 2, (4 − 2 − 2)/(2 − 2) = 0/0 — indeterminate, so factor.
- +1(a) Factor & cancel: x² − x − 2 = (x − 2)(x + 1); for x ≠ 2 this is x + 1.
- +1(a) Substitute now: limx→2 (x + 1) = 3.
- +1(b) Bound the wild factor: −1 ≤ sin(1/x) ≤ 1, so −x² ≤ x² sin(1/x) ≤ x².
- +1(b) Squeeze: both −x² → 0 and x² → 0 as x → 0, so the middle is trapped at 0.
- +1(b) Conclude: limx→0 x² sin(1/x) = 0, even though sin(1/x) itself has no limit.
Key terms
- Limit
- The value f(x) approaches as x approaches a, written limx→a f(x). It is determined by behaviour near a, not the value at a — so a limit can exist where f is undefined.
- Indeterminate form (0/0)
- A quotient that evaluates to 0/0 on substitution. It does not mean the limit fails to exist; it signals that algebra (factor and cancel, or a conjugate) or L'Hôpital is needed to resolve it.
- One-sided limit
- The value approached from one direction only: the left limit (x → a−) and the right limit (x → a+). The two-sided limit exists only when both one-sided limits exist and are equal.
- Squeeze (sandwich) theorem
- If g(x) ≤ f(x) ≤ h(x) near a and lim g = lim h = L, then lim f = L. It is the standard way to evaluate limits of bounded-times-vanishing products such as x² sin(1/x).
- Continuity
- f is continuous at a when limx→a f(x) = f(a): no hole, no jump, no blow-up. Continuity is what licenses direct substitution and is the hypothesis behind the Intermediate Value Theorem.
Limits and Continuity FAQ
Why can't I just substitute x = a straight away?
You can — but only when the function is continuous at a. If substitution gives a defined value, that value is the limit. If it gives an indeterminate form like 0/0, substitution is illegal at that point and you must do algebra first (factor and cancel, or rationalise a conjugate). A limit cares about x near a, not at a, which is exactly why the cancelled factor doesn't matter.
When does a two-sided limit fail to exist?
When the left and right one-sided limits disagree (a jump), when the function blows up to ±∞, or when it oscillates without settling (like sin(1/x) near 0). Always check both sides at a point where the rule for f changes — piecewise functions and absolute values are the classic traps.
How do I know to use the squeeze theorem?
Look for a bounded factor multiplied by something that vanishes: sin or cos of a wild argument (which always sits in [−1, 1]) times a factor going to 0. Bound the wild factor between −1 and 1, multiply through by the vanishing factor, and show both outer bounds share the same limit. The middle is then trapped at that limit.
What is the Intermediate Value Theorem actually for?
It guarantees a root or a target value exists without finding it. If f is continuous on [a, b] and a value N lies between f(a) and f(b), then f hits N somewhere in (a, b). In exams it is used to prove an equation has a solution: show f is continuous and that it changes sign across an interval, and the IVT does the rest.
Exam move
Build a reflex for the first decision: substitute, and read the result. A defined number is the answer; a 0/0 form means factor-and-cancel or a conjugate; a bounded-times-vanishing product means squeeze. Memorise the standard limits (especially sin(x)/x → 1) so you recognise them inside bigger expressions. For one-sided work, always test both sides at a rule change or an absolute value. Reserve the IVT for existence arguments — state continuity, show a sign change, conclude a root exists. Because Quiz A is no-calculator, keep every value exact and show the cancellation line; the method marks are awarded for the algebra, not just the final number.
