A ball rolls down a ramp, speeding up the whole way. Its position after \(t\) seconds is \(p(t)=t^2\) meters. Easy question: what's its average speed between \(t=3\) and \(t=4\)? Distance over time: \(\tfrac{16-9}{1}=7\) m/s. Hard question: how fast is it going at the exact instant \(t=3\)? Speed needs two moments to compare — and an instant only has one. Divide by zero time and the formula collapses. Watch what happens if we don't give up, and instead shrink the window:
What you're looking at: the curve is the ball's position over time. The straight line connects two moments — its slope is the average speed between them. The window between the two moments is about to shrink.
| window | avg speed (m/s) |
|---|
The window keeps shrinking, the averages keep settling: 7, 6.5, 6.1, 6.01, 6.001… They're heading somewhere. No single window ever gives exactly 6 — but every shrink gets closer. The speed at the instant is the number the averages approach: 6 m/s. That move — answering an impossible question by approaching it — is the limit.
Here's the same idea on a stranger function: \(f(x)=\dfrac{x^2-1}{x-1}\). Try plugging in \(x=1\): the top is \(0\) and the bottom is \(0\) — division by zero, so \(f(1)\) simply does not exist. The graph has a hole there (drawn as an open circle). But the limit question isn't about \(x=1\) itself — it's about the approach. Send walkers in from both sides and watch where they're heading:
What you're looking at: two dots ride the curve — one coming from the left of \(x=1\), one from the right. The open circle at the top of their path is the hole: the one point the function skips.
| x (from left) | f(x) | x (from right) | f(x) |
|---|
Both walkers close in on height 2 — even though the function never actually gets there. The limit is where the function is heading, not where it is. Say that sentence out loud; it's the most important one in the unit.
You just computed your first limit by watching it happen. Here's how mathematicians write what you saw:
The walkers came from two directions. Usually they agree on a destination — but not always. Here's a function with a jump: it follows one rule before \(x=2\) and a different rule after. Send in the walkers one side at a time:
What you're looking at: the left branch ends in an open circle at height 2 (the function gets arbitrarily close to that point but it's not included); the right branch starts with a filled dot at height 3 (that point IS the function's value at \(x=2\)).
The little superscript minus and plus mark the direction: \(x\to 2^-\) approaches from below (the left), \(x\to 2^+\) from above (the right). The two-sided limit exists only when both one-sided limits exist AND agree. Here they don't — so the limit does not exist (written DNE), even though \(f(2)\) is perfectly fine.
This is the AP exam's favorite opening move: show a graph, then ask about the limit, the function's value, and whether they match — three separate questions with possibly three different answers. Five rounds. Read each graph carefully; open circle = not included, filled dot = the actual value.
Legend, every round: an open circle ⭘ marks a point the graph approaches but does NOT include; a filled dot ⬤ marks the function's actual value.
No graph handy? A table of values can show you where a function is heading. Plug in inputs marching toward the target and watch the outputs settle. But there's an honest warning attached — tables can lie. Both stories, side by side:
| x | f(x) |
|---|
The first table is honest: the outputs settle on 0.25 and the limit really is \(\tfrac14\). The second is a trap — the outputs LOOK like they're settling on 0, but the chosen x-values just happened to land on the function's zero-crossings. Tables suggest; they never prove.
The limit is the destination of the approach — the function's value at the point (or whether it even has one) is a completely separate fact.
A two-sided limit is a unanimous verdict: left walker and right walker must arrive at the same height, or the answer is DNE.
\(x\to a\) means arbitrarily close, never equal. That's why a hole at the target changes nothing — the walkers never stand on it anyway.
Questions that collapse at a point (speed at an instant, slope at a point) get answered by approaching the point — the move the entire rest of calculus is built on.
Five problems, ramping from reading to reasoning. Try each before revealing.
Eight fast checks across the whole lesson.
The single most common Unit 1 error. The filled dot is f(a); the limit is where the curve is heading. On a graph question, locate the approach FIRST, the dot second.
Backwards — a hole at the target is exactly the situation limits were built for. The approach machine's limit was a clean 2 with no f(1) at all.
A two-sided limit needs BOTH one-sided limits, agreeing. Piecewise functions and absolute values are the classic places the sides quietly differ.
\(x\to 2\) is an approach, not a substitution command. (Substituting often works — Lesson 2 explains exactly when — but the MEANING is the approach.)
A table is a handful of samples. The sin(π/x) trap looked like it settled on 0 while actually oscillating forever. Tables estimate; algebra and theorems prove.
DNE is a precise verdict with specific causes: sides disagree, values grow without bound, or endless oscillation. A hole alone is NOT one of them.