Where a quantity is heading can be perfectly clear — even when it never arrives.
Some of the most useful numbers in mathematics are ones a quantity never quite reaches — it only approaches them. A sequence of guesses can crowd onto a single value; a curve can climb toward a height it never lands on; a speed can be defined at one instant by squeezing an average over shorter and shorter stretches. The tool that captures all of this is the limit: the value a quantity heads toward as its input creeps toward a target. In this lesson we make "closer and closer" precise, watch a sequence and then a function approach a limit from both sides, learn the simple arithmetic that lets limits add and multiply, and finally meet continuity — the clean case where a function's limit and its actual value agree, so you never lift your pen. This is the first idea of calculus, and everything in Stage 36 grows from it.
Picture walking toward a wall, each step covering half the remaining gap. You get closer and closer — yet you never actually touch it. "Closer and closer" is a feeling; mathematics needs a sentence that means it exactly. That sentence is the limit.
When we write the arrow x → a we mean: let x creep toward the target a, from either side, getting as close as we like — without insisting it ever reach a. And
limx→a f(x) = L
reads "as x heads to a, the height f(x) heads to L." The number L is the destination the outputs are aiming at. Crucially, this says nothing about the value at a itself — f might be undefined there, or sit somewhere else entirely. A limit is a statement about the journey, not the arrival.
A limit describes where the outputs are heading as the input approaches a target. The arrow x → a means "x gets arbitrarily close to a" — it never requires x = a, and the limit can exist even where the function does not.
The cleanest first example is a sequence — a list of stepping-stone numbers a1, a2, a3, … indexed by the counting numbers (you met these in Stage 26). Take
an = 1 + 1n
The terms are 2, 1.5, 1.333…, 1.25, 1.2, … The extra bit 1/n shrinks toward 0, so the whole term presses down onto 1. We never hit 1 (the gap is always 1/n, never zero), but we get as close as we please. We write limn→∞ an = 1: the stepping stones converge on one spot.
However far out you push N, the last few terms sit a hair above the line and that hair keeps thinning. That is convergence: for any tolerance you name, all terms past some point land inside it.
Now do the same for a curve. To find limx→a f(x), walk in toward x = a along the graph — once from the left and once from the right — and watch the height. If both approaches head to the same number, that number is the limit.
The famous case is the curve with a hole:
f(x) = x² − 1x − 1
At x = 1 this is 0/0 — undefined, a genuine gap. But for every other x we can factor and cancel: (x²−1)/(x−1) = (x−1)(x+1)/(x−1) = x + 1. So away from 1 the curve is just the line y = x + 1, which heads to 1 + 1 = 2 as x → 1. Both sides agree, so limx→1 f(x) = 2 — even though f(1) does not exist.
Find limx→3 (x² − 9)/(x − 3). The plug-in gives 0/0, so factor: (x−3)(x+3)/(x−3) = x + 3 for x ≠ 3. As x → 3 this heads to 3 + 3 = 6. The limit is 6, and the point at x = 3 is a removable hole.
Most limits don't need a clever trick — they split into easy pieces. If limx→a f = F and limx→a g = G, then limits respect the four operations:
| operation | limit of the combination |
|---|---|
| sum | lim (f + g) = F + G |
| difference | lim (f − g) = F − G |
| product | lim (f · g) = F · G |
| quotient | lim (f / g) = F / G (needs G ≠ 0) |
So limx→2 (x² + 3x) is just 4 + 6 = 10 — break it into lim x² and lim 3x and add. The only catch is the quotient law's fine print: the bottom limit must not be 0. When it is 0 (and the top is too), you get the 0/0 "indeterminate" form — the symbol alone can't decide the answer, and you must do extra work (factor, cancel, simplify) to see where things really head. A celebrated example is limx→0 (sin x)/x = 1: plugging in gives 0/0, yet the true destination is exactly 1.
A limit is about the approach, not the value at a. A function can have a limit at a point where it is undefined (our hole, where lim = 2 but f(1) does not exist), or where f(a) sits somewhere else. "Just plug in x = a" works only when the function is nice there — and "x actually reaches a" is the wrong mental picture.
The happy case is when nothing surprising happens at a — the limit and the actual value agree. We say f is continuous at a when
limx→a f(x) = f(a)
three things at once: the limit exists, f(a) exists, and they are equal. Geometrically, you can draw the curve through x = a without lifting your pen. Polynomials are continuous everywhere; so are sines, cosines, and exponentials. When continuity fails, the break comes in one of three flavors:
Is f(x) = (x² − 1)/(x − 1) continuous at x = 1? The limit exists (2), but f(1) does not exist, so the three-part test fails — it is not continuous there. The break is removable: define f(1) = 2 and the patched function is continuous.
| idea | what it says |
|---|---|
| limit | limx→a f(x) = L: outputs head to L as x approaches a (from both sides) |
| sequence limit | an = 1 + 1/n → 1: terms crowd onto one value |
| two-sided | the limit exists only if left = right; otherwise it doesn't exist |
| arithmetic | limits add, subtract, multiply, divide (quotient needs bottom ≠ 0) |
| 0/0 | indeterminate — factor & cancel to find the true destination |
| continuity | limx→a f(x) = f(a): never lift the pen |
| breaks | jump · removable · infinite |
Write the first five terms of an = 1 + 1/n and state limn→∞ an.
2, 1.5, 1.333…, 1.25, 1.2. The 1/n part shrinks to 0, so the terms head to 1: lim = 1.
Evaluate limx→2 (x² + 3x − 1) using the limit laws.
Polynomials are continuous, so just substitute: 2² + 3·2 − 1 = 4 + 6 − 1 = 9.
Evaluate limx→4 (x² − 16)/(x − 4). (Plugging in gives 0/0.)
Factor: (x−4)(x+4)/(x−4) = x + 4 for x ≠ 4. As x → 4 this heads to 4 + 4 = 8. The point at x = 4 is a removable hole.
A function approaches height 3 from the left of x = 1 and height 5 from the right. Does limx→1 f(x) exist? What kind of break is this?
No — the two sides disagree (3 ≠ 5), so the limit does not exist. This is a jump discontinuity.
For g(x) = (x² − 9)/(x − 3) we found limx→3 g = 6, but g(3) is undefined. How would you make g continuous at 3, and what break was it?
Define g(3) = 6 (the limit value); the patched function is continuous. The break was removable.
True or false: if f(a) is undefined, then limx→a f(x) cannot exist. Explain.
False. A limit depends only on the approach, not the value at a. Our hole example has limx→1 f = 2 even though f(1) does not exist.
Six questions to lock it in. Tap the answer you think is right.
This lesson opens the bridge into calculus by formalizing the intuition of "approaching." It supports HSF-IF.B.6 (interpreting and estimating rates of change as a foundation for limits) and lays the precalculus groundwork for limits and continuity that AP Calculus AB/BC builds on directly. The two-sided-approach and continuity-classification activities make the formal definition concrete before any symbolic machinery, and the deliberate "limit vs. value at a point" distinction heads off the most common early misconception. A good follow-up is asking a student to argue, in words, why a function can have a limit exactly where it is undefined.