MAST10006 · Calculus 2
Limits and Continuity
A limit describes where a function is heading as the input closes in on a point — whether or not the function is even defined there. It is the engine under derivatives, continuity, series and every rate in the rest of Calculus 2, and the exam tests it three ways: evaluate the limit, justify it with a named law or theorem, and decide continuity — all by hand. This opening chapter sets up the limit laws (which need both pieces to exist before you split a product or quotient), the handful of standard limits worth memorising, the squeeze (sandwich) theorem for functions that oscillate too violently to attack directly, the three-condition definition of continuity, and L'Hopital's rule for the indeterminate forms 0/0 and ∞/∞. The recurring marked discipline is to substitute first, name the indeterminate form, and only then differentiate.
What this chapter covers
- 011.1 The working definition — one-sided limits and existence
- 021.2 The limit laws (and why both pieces must exist first)
- 03Standard limits you must know cold
- 04The squeeze (sandwich) theorem
- 05Continuity — the three conditions and the failure modes
- 06L'Hopital's rule — verify the form before differentiating
Worked example: a limit by manufacturing a standard form
- +1Rewrite tan: tan 2x = sin 2x / cos 2x, so the expression is (sin 3x / sin 2x) · cos 2x.
- +1Manufacture the standard forms. Multiply top and bottom to expose sin(□)/□: (sin 3x / 3x) · (2x / sin 2x) · (3x / 2x) · cos 2x.
- +1Take each factor's limit as x → 0: sin 3x / 3x → 1, 2x / sin 2x → 1, 3x / 2x = 3/2, cos 2x → 1. Each factor converges, so the product law applies.
- +1Multiply: 1 · 1 · (3/2) · 1 = 3/2.
Key terms
- One-sided limit
- The value a function approaches from a single side of a point. The two-sided limit limx→a f(x) exists if and only if the left limit and the right limit both exist and are equal — the value at a itself is irrelevant.
- Squeeze theorem
- If g(x) ≤ f(x) ≤ h(x) near a and g and h both tend to the same L, then f is forced to L as well. The go-to tool when an oscillating factor (such as sin(1/x)) prevents a direct attack; the two bounds must tend to the SAME, KNOWN limit.
- Continuity
- f is continuous at a when three conditions hold together: f(a) exists, the two-sided limit exists, and the limit equals the value. Failure modes are removable (a hole), jump (one-sided limits differ) and infinite (a vertical asymptote).
- Indeterminate form
- An expression such as 0/0 or ∞/∞ whose limit cannot be read off by substitution — it could be anything until resolved. L'Hopital's rule applies only after you have confirmed the form is genuinely indeterminate.
- L'Hopital's rule
- For a 0/0 or ∞/∞ quotient, lim f/g = lim f′/g′ provided the right-hand limit exists. You may iterate while each new quotient stays indeterminate; applying it to a finite/non-zero form is a marked error.
Limits and Continuity FAQ
Does a limit existing mean the function is defined there?
No — this is the central distinction of the chapter. A limit watches what happens near a, never at a, so it can exist where f(a) is undefined (a hole), and conversely f(a) can exist but differ from the limit. Sorting out exactly when limit equals value is the whole content of continuity.
When do I use the squeeze theorem instead of just substituting?
Reach for it when a factor oscillates so violently you cannot evaluate the limit directly — the classic case is x² sin(1/x). Bound the wild factor by constants (almost always −1 ≤ sin(·) ≤ 1), multiply through by the tame factor (mind the sign), check both outer bounds tend to the same L, then conclude. If the bounds tend to different values the theorem fails.
What is the most common L'Hopital mistake?
Skipping the verification line. Applying L'Hopital to a form that is not 0/0 or ∞/∞ is a marked error and usually gives the wrong answer. Always substitute first, state the form explicitly in your working, and only then differentiate — and stop iterating once the quotient is no longer indeterminate.
How do I make a piecewise function continuous at a join?
Polynomials, sin, cos, ex and log are continuous on their domains, so a piecewise function can only break at the boundary points. Match the two one-sided limits to the assigned value there: continuity needs limit = value. For differentiability you must additionally match the one-sided derivatives, since differentiable implies continuous but not the reverse (|x| at 0 is the counterexample).
Exam move
Run a fixed routine on every limit question. First substitute to see what you have; if it is a clean number you are done, and if it is 0/0 or ∞/∞ name the form out loud on the page before doing anything else. For the standard limits (sin x / x → 1, (ex − 1)/x → 1) the move is to manufacture the known shape by multiplying and dividing, then let each block go to 1. For oscillating factors, reach for the squeeze theorem and check both bounds tend to the same value. For continuity, check the three conditions one at a time at the joins only. Marks are lost not on the final number but on the missing justification — name the law or theorem, and confirm its hypotheses (both limits exist, the form really is indeterminate, the bounds really agree) every time.
