How to Find Limits: Every Method, in Order

The order to try methods

The fastest way to find a limit is to substitute the value $x$ approaches. If the function is defined and continuous there, the number you get is the limit, and you are done (Topic 1.5). Everything else in this guide exists for the cases where substitution fails. So the real skill is not memorizing five techniques: it is reading the form substitution hands back and letting that form pick the technique.

After you substitute, one of a few things happens. Match what you see to the row below, then jump to that method. This table is the whole decision procedure, and the rest of the guide just fills in each row.

Direct substitution: the default

Substitution works because of the limit laws and continuity. The limit laws (Topic 1.5) say the limit of a sum, difference, product, quotient, or composition is built from the limits of the pieces, and polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous everywhere they are defined (Topic 1.12). Put those together and, for any of those functions, the limit as $x \to c$ equals the value at $c$ whenever $c$ is in the domain (continuity at a point, Topic 1.11).

So $\lim_{x \to 4} (x^2 - 3x + 1)$ is just $16 - 12 + 1 = 5$. No manipulation, no special cases. If you can plug in the number and nothing blows up, that is the whole problem. Reach for the harder methods only when substitution produces $\frac{0}{0}$.

This is why so many textbook limits look trivial: they are continuity in disguise. The interesting problems are engineered so that substitution breaks, which is the exam's way of testing whether you can recognize and repair the break. When it breaks, it almost always breaks as $\frac{0}{0}$, and that specific failure is what the next two sections handle.

When substitution gives 0/0: factor or use the conjugate

If substituting $c$ makes both the numerator and denominator zero, a common factor of $(x - c)$ is hiding. For polynomials, $(x - c)$ divides both of them directly; when a radical is present, that shared $(x - c)$ factor is exposed only after you multiply by the conjugate. Either way, that shared factor is what makes the fraction look like $\frac{0}{0}$, and your job is to expose it and cancel it (Topic 1.6).

Factoring is the move when you have polynomials. Factor the top and bottom, cancel the common $(x - c)$, and substitute into what is left. The cancellation is legal because a limit only cares about $x$ near $c$, not at $c$, so dividing by $(x - c)$ is fine for every $x$ the limit actually looks at.

The conjugate is the move when a square root sits inside the $\frac{0}{0}$ expression. Multiply the numerator and denominator by the conjugate of the radical part, which turns a difference like $\sqrt{x+4} - 2$ into a difference of squares. The radical disappears, a factor cancels, and substitution finishes the job. Quick recognition rule: polynomials on both sides means factor; a stubborn square root means conjugate.

Trig limits and the squeeze theorem

Some $\frac{0}{0}$ limits will not factor and have no radical to rationalize, because the trouble is trigonometric. Two results, both provable with the squeeze theorem, cover almost every case on the exam:

When you see a sine divided by its own argument (or something you can reshape into that), match it to the first result. For example, $\lim_{x \to 0} \frac{\sin 3x}{4x}$ is $\frac{0}{0}$, but rewriting gives $\frac{\sin 3x}{4x} = \frac{3}{4} \cdot \frac{\sin 3x}{3x}$. As $x \to 0$ the argument $3x \to 0$ as well, so $\frac{\sin 3x}{3x} \to 1$ and the limit is $\frac{3}{4}$.

The squeeze theorem is the tool of last resort, and the one to reach for when a function oscillates. It says that if $g(x) \le f(x) \le h(x)$ for all $x$ near $c$, and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$ too. Use it when substitution and algebra both fail because a bounded but wildly oscillating piece like $\sin\!\left(\frac{1}{x}\right)$ refuses to settle, yet you can trap the whole expression between two functions that meet at the same value (Topic 1.8).

Pitfalls and how to choose fast

• $\frac{0}{0}$ versus $\frac{c}{0}$: a zero over a zero is indeterminate and calls for algebra; a nonzero number over zero is a vertical asymptote, so the limit is infinite or does not exist. They look similar and lead to opposite conclusions.
• Do not stop at $\frac{0}{0}$ and write 'does not exist.' That form means the answer is still reachable, usually with one factoring or conjugate step.
• Check one-sided limits when the function is piecewise, or when $x$ approaches the edge of a domain or a vertical asymptote. If the left and right limits disagree, the two-sided limit does not exist (Topic 1.3).
• L'Hôpital's Rule (BC only, Topic 4.7) is not needed for anything in Unit 1. Every $\frac{0}{0}$ here yields to factoring, a conjugate, or a trig limit.
• If a function oscillates without settling, either squeeze it or conclude the limit does not exist. Oscillation never factors away.

Run the same loop every time: substitute, read the form, and let the form choose the method. That single habit, not raw memorization, is what makes limit problems fast.