Limits & Continuity, taught from zero

1 · The problem limits were invented to solve

Take this function and try to evaluate it at x = 3:

Plugging in gives 0/0 — meaningless. The function is simply not defined at 3. And yet something interesting is happening there. Feed it inputs near 3 and watch:

The outputs are closing in on 6 from both sides, and nothing can stop them. The function has a hole at x = 3, but the hole is clearly "at height 6". That number — the value the outputs settle toward — is the limit:

Read it as: "as x approaches 3, f(x) approaches 6." Two things to notice, because they are the whole concept:

The limit is about the journey, not the destination. We never plug in 3. We only ask what the outputs are heading toward. Whether f(3) exists, doesn't exist, or equals something weird like 42 — the limit doesn't care. It describes the approach.

Both sides must agree. We came from the left (2.9, 2.99, …) and from the right (3.1, 3.01, …) and got the same answer. That agreement is part of the definition — hold that thought for section 4.

Why does this matter? Because all of calculus is built on it. A derivative is a limit (the slope of a chord as the two points merge). An integral is a limit (a sum of slivers as they shrink). This chapter is the foundation the rest of the course stands on — that's why the course wants it solid before anything else.

2 · When plugging in just works

Here is the anticlimax: for most functions in this course, the limit at a point is found by… plugging the point in. lim_x→2 (x² + 1) = 5. Done. Polynomials, sin, cos, e^x, and their sums, products and compositions are all "continuous" (section 5 makes that precise), which means their limits are just their values.

So when is a limit problem actually a problem? Almost always when plugging in gives 0/0. That form is a flashing sign that says: something is hiding here — dig. The whole skill of §2.4 is a small toolbox for digging. There are exactly three tools.

3 · The 0/0 toolbox — three tools, used forever

Tool 1: Factor and cancel

You already used it in the check above. A 0/0 from polynomials means top and bottom share a factor that is becoming zero — find it, cancel it, plug in what's left:

One conceptual point worth a pause: cancelling is legal precisely because limits ignore the point itself. The functions (x²−9)/(x−3) and x+3 differ only at x = 3 — and the limit never looks there.

Tool 2: Rationalise (the conjugate trick)

When the 0/0 involves a square root, factoring won't work directly. The move is to multiply top and bottom by the conjugate — the same expression with the middle sign flipped. Why? Because (A − B)(A + B) = A² − B²: the multiplication squares the root away, and the hidden common factor pops out. Watch it happen:

The numerator became (x+1) − 1 = x, which cancelled the x below — that x was the hidden shared factor all along. After cancelling, plugging in x = 0 is safe.

Tool 3: The one limit you memorise

Here is why it's believable. For a tiny angle, the arc of the unit circle it cuts and the vertical chord it raises are almost the same length — bend a 1 mm piece of wire into an arc and it's still essentially 1 mm tall. The arc length is θ (that's what radians mean), and the height is sin θ, so their ratio squeezes to 1. Try it: sin(0.1) = 0.0998… — already within 0.2% of 0.1.

This is also why the "radians" warning is not pedantry: in degrees, the arc cut by 1° is tiny while "1" is not, and the ratio heads to π/180 instead of 1. Every trig-limit formula in this course silently assumes radians.

Bonus tool: the Sandwich (Squeeze) Theorem

Sometimes a function is too wild to compute directly but is trapped between two tame ones. If g(x) ≤ f(x) ≤ h(x) near a point and the outer two both approach the same limit L, then f has nowhere else to go: it approaches L too. Think of two escorts walking a prisoner — where they converge, he goes.

The signature example: lim_x→0 x² sin(1/x). As x → 0, sin(1/x) oscillates faster and faster — it has no limit of its own. But it never leaves [−1, 1], so the whole thing is trapped: −x² ≤ x² sin(1/x) ≤ x². Both fences collapse to 0, so the limit is 0. The wildness got squeezed out.

4 · One-sided limits — when the two directions disagree §2.4

So far the left and right approaches always agreed. They don't have to. The cleanest troublemaker is:

For any positive x, |x| = x, so f(x) = 1. For any negative x, |x| = −x, so f(x) = −1. The graph is two flat shelves with a cliff at 0. Approach 0 from the right and the outputs sit at 1 the whole way; from the left, −1. We write:

(The little ⁺ means "from the right", ⁻ "from the left".) And the rule that ties it together: the two-sided limit exists only if both one-sided limits exist and are equal. Here they're not, so lim_x→0 |x|/x simply does not exist — and "does not exist" is a legitimate, full-credit answer when it's true.

Where you'll actually use this: piecewise functions — functions defined by different formulas on different intervals. At the "seam" where the formula changes, the only way to know what's happening is to compute the left limit from one formula and the right limit from the other, and compare. Exams love this; section 5's check gives you one.

5 · Continuity — when the graph has no surprises §2.5

Informally, a function is continuous at a point if you can draw the graph through it without lifting your pen. Formally, three things must hold at x = a, and each one can fail on its own:

1. f(a) exists — the point is actually on the graph (fails: a hole),
2. lim_x→a f(x) exists — both sides agree on where the graph is heading (fails: a jump, like |x|/x),
3. the two are equal — the graph arrives where the point actually is (fails: a hole with the dot pasted somewhere else).

Exam answers are expected to check all three, in order — it takes two lines and reads as rigour.

The payoff of knowing a function is continuous is the anticlimax from section 2: limits become substitution. Polynomials, sin, cos, e^x are continuous everywhere; rational functions everywhere their denominator isn't zero; and sums, products, quotients and compositions of continuous functions stay continuous. That one paragraph disposes of 90% of "is it continuous?" questions.

The Intermediate Value Theorem, in one metaphor

If you walk continuously from the south bank of a river to the north bank, at some moment you were in the river. That's the IVT: a function continuous on [a, b] takes every value between f(a) and f(b) somewhere in between. No skipping.

Its standard exam use is locating roots without solving: to show x³ + x − 3 = 0 has a solution in [1, 2], compute f(1) = −1 (below zero) and f(2) = 7 (above zero), note the function is a polynomial hence continuous, and cite IVT: somewhere in between it crossed zero. Three lines, full marks.

6 · The precise definition: ε–δ as a game §2.3

Everything so far said "gets close to" — which is honest but vague. How close is close? §2.3 replaces the vagueness with a challenge game, and this part is new to everyone, BITSAT background or not — 12th boards never make limits rigorous.

Claiming lim_x→a f(x) = L means claiming you can win the following game every time:

The skeptic moves first: "I don't believe the outputs really settle at L. Get them within ε = 0.01 of L." (ε is the skeptic's tolerance — they can make it as brutal as they like.)

You answer with a δ: "Keep x within δ = 0.003 of a, and I guarantee every output lands within your 0.01 of L."

If you have a winning answer for every possible ε — not just 0.01, but 0.0001 and a billionth — then the limit is L. Written formally, that's exactly the sentence in the textbook:

Decode it symbol by symbol: |x − a| < δ is "x within δ of a"; the 0 < in front excludes x = a itself (journey, not destination); |f(x) − L| < ε is "output within ε of L"; and "for every ε… there is a δ" is the game: skeptic first, you second. There is an interactive version of this game on the notes page — go slide the ε and watch the δ respond, then come back.

How do you actually find the winning δ? For the linear functions this course examines, it's two lines of algebra, always the same two lines. Prove lim_x→2 (3x − 1) = 5:

1. Write down the thing that must be small, and factor the distance to a out of it: |f(x) − 5| = |3x − 6| = 3|x − 2|.
2. You need 3|x − 2| < ε. That happens exactly when |x − 2| < ε/3. So answer the skeptic with δ = ε/3, and the game is won — for every ε at once, since the recipe works whatever ε they name.

Notice δ came out as ε divided by the slope, 3. That's not a coincidence: a line with slope 3 stretches horizontal distances into vertical ones 3× as big, so to land within ε vertically you must start within ε/3 horizontally. Steeper line, smaller δ. Once you see that, every linear ε–δ proof is the same proof.

7 · Limits at infinity, and asymptotes §2.6

Last question of the block: what happens to a function as x marches off to ±∞? For the rational functions (polynomial over polynomial) this course cares about, it's a tug of war between the top and bottom — and the highest power wins. Everything else becomes irrelevant at scale: when x is a million, 3x² dwarfs any 5x or constant riding along.

Three cases, by comparing the top degree with the bottom degree:

• Degrees equal → the leading coefficients fight it out. (3x²+5)/(2x²−x) → 3/2. The graph flattens onto the horizontal line y = 3/2 — a horizontal asymptote.
• Bottom degree higher → the bottom wins → limit 0. The x-axis is the asymptote.
• Top degree higher → the top wins → the function blows up to ±∞. No horizontal asymptote. (If the top is exactly one degree higher, the graph hugs a slanted line — an oblique asymptote, found by polynomial long division; the notes page unpacks one.)

And the other kind of asymptote: vertical, where the function blows up at a finite x — wherever the denominator hits 0 while the numerator doesn't. The italics carry a trap: if top and bottom are both zero there, you're back in 0/0 territory — factor first, because it may cancel into a mere hole.

8 · You're ready — what to do next

You now hold the complete conceptual kit for §2.3–2.6: what a limit is, the three 0/0 tools, one-sided limits, the three-part continuity check, IVT, the ε–δ game, and the asymptote rules. Two follow-ups turn it into marks:

• Drill it: the notes page has the section-by-section exercise table for Thomas (odd numbers — answers in the back). The lesson gave you understanding; the exercises make it fast.
• Revise from the notes page, not this one. This page is deliberately slow; before a quiz you want the compressed map.