Why does plugging in usually work? Because most functions are continuous — the word from Lesson 1's quiz, meaning the graph flows through the point with no hole or jump, so where the function is heading is exactly where it is. Polynomials, and quotients where the bottom isn't zero, are continuous everywhere they're defined. For those, the limit is pure substitution. Try it:
The first three substitute cleanly — green light, you're done. The fourth plugs in to give \(\tfrac00\): a red flag that there's digging to do. That's the whole fork in the road.
Substitution works because limits play nicely with every basic operation. If \(f\) and \(g\) each have a limit as \(x\to a\), then the limit of their sum is the sum of their limits, and the same for differences, products, and (if the bottom isn't zero) quotients. In plain terms: you can break a complicated limit into the limits of its pieces.
So you substitute, and you get \(\tfrac00\). It's tempting to write "undefined" and move on — but that's the wrong instinct. \(\tfrac00\) is called an indeterminate form, and the name is the whole point: it means the answer is not yet determined, not that there's no answer. Compare:
\(\tfrac00\) means a common factor of \((x-a)\) is hiding in both the top and bottom. Cancel it and the hole disappears, revealing the limit. The next two tools are just two ways to do that cancelling.
When the top and bottom are polynomials and you hit \(\tfrac00\), the cure is factoring: the thing making both zero is a shared \((x-a)\) factor. Cancel it — legally, because \(x\to a\) means \(x\ne a\), so you're never dividing by zero — and substitute into what survives.
What you're looking at: the graph has an open hole where substitution gave 0/0. Step through — when the \((x-a)\) cancels, watch the hole fill in.
Each one substitutes to \(\tfrac00\) first — that's your cue to factor. Watch the \((x-a)\) appear in both top and bottom, then vanish.
Factoring needs polynomials. But what about that mystery from Lesson 1's table, \(\dfrac{\sqrt{x+4}-2}{x}\) — the one that seemed to head toward \(\tfrac14\)? A square root blocks ordinary factoring. The trick: multiply top and bottom by the conjugate (the same expression with the sign flipped). The difference-of-squares pattern \((\sqrt{A}-B)(\sqrt{A}+B)=A-B^2\) erases the radical, and the troublesome factor finally cancels.
What you're looking at: same idea as factoring — the radical hides a removable hole. When the troublesome factor cancels, the hole fills.
The first one is the table mystery from Lesson 1, now solved exactly: the answer really is \(\tfrac14\), no table needed. The conjugate turns "I think it's heading there" into "it is."
This is the skill the AP exam tests directly: not doing a technique, but choosing it. The routine is always the same three moves — and it starts with substitution every single time. Diagnose each limit below, then pick the right first step:
Every limit computation begins by plugging in. Most of the time you're instantly done; the rest of the time, what you get tells you exactly what to do next.
The indeterminate form is a promise that an answer exists once you cancel the shared \((x-a)\). Never stop at \(\tfrac00\).
Factoring and the conjugate both do the same thing — expose and cancel the hidden factor. Polynomials factor; radicals need the conjugate.
A nonzero number over zero isn't indeterminate — it blows up. Don't waste time factoring it; it's heading to infinity or DNE.
Five problems spanning all three tools — including a complex fraction, a notorious exam favorite. Try each before revealing.
Eight fast checks across the whole lesson.
The single biggest error of the lesson. \(\tfrac00\) is indeterminate — it's the START of the work, not the end. There's a hidden factor; go find it.
Only \(\tfrac00\) is indeterminate. \(\tfrac50\) genuinely blows up — factoring it is wasted effort. Check WHICH zero-over-zero you actually have.
After cancelling, you're not done — plug \(a\) into the simplified expression. The cancel reveals the function; substitution finishes the limit.
The conjugate flips the sign between the two terms of the radical expression — \(\sqrt{A}-B\) pairs with \(\sqrt{A}+B\). Multiply BOTH top and bottom by it, or you've changed the problem.
Substitute first. If it gives a clean number, there's nothing to cancel and no reason to factor — you're already done.
Cancelling \(\tfrac{x-a}{x-a}\) is legal precisely because \(x\to a\) means \(x\ne a\). The limit lives in the approach, where that factor is genuinely nonzero.