The big idea

A limit asks: "As $x$ gets arbitrarily close to $a$, what value is $f(x)$ getting arbitrarily close to?"

It does not ask what $f(a)$ is. That's the trick: $f(a)$ may be undefined, the wrong value, or anything else, and the limit can still exist. This separation between value at a point and behavior near a point is exactly what lets calculus describe instantaneous rates and exact areas.

Three things a limit can do at $x = a$

1. Equal a finite number. $\lim_{x\to a} f(x) = L$. The function approaches a single value from both sides.
2. Equal $\pm\infty$ (an "infinite limit"). The function blows up; this signals a vertical asymptote at $x = a$.
3. Fail to exist. Either the left- and right-hand limits disagree (jump), the function oscillates without settling, or $\pm\infty$ disagrees on the two sides.

A limit exists (in the AP sense of "equals a number") only when both one-sided limits agree on a finite value. Any other behavior — jumps, oscillation, blow-ups — means the limit DNE.

Computing limits — the AP playbook

When asked to evaluate $\lim_{x \to a} f(x)$, follow this order:

1. Try direct substitution. If $f$ is continuous at $a$, $\lim = f(a)$. Done.
2. If you get $0/0$, look for algebraic simplification. Most common moves:
   • Factor and cancel (e.g., $\frac{x^{2} - 4}{x - 2} = x + 2$ for $x \neq 2$).
   • Rationalize (multiply by conjugate when there's a square root).
   • Combine fractions in the numerator.
   • Use a trig identity (e.g., $\sin^{2}x + \cos^{2}x = 1$).
3. Recognize special trig limits: $\lim_{x\to 0} \frac{\sin x}{x} = 1$ and .
4. For limits at infinity of rational functions, compare leading-term degrees:
   • degree(num) < degree(den): limit = 0
   • degree(num) = degree(den): limit = ratio of leading coefficients
   • degree(num) > degree(den): limit is $\pm\infty$ (no horizontal asymptote)
5. If you still get an indeterminate form ($0/0$ or $\infty/\infty$), use L'Hôpital's Rule: $\lim \frac{f}{g} = \lim \frac{f'}{g'}$.

Continuity in one breath

A function $f$ is continuous at $x = a$ iff all three are true:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

If any one fails, $f$ is discontinuous at $a$. The flavor depends on which one:

The Intermediate Value Theorem (IVT)

If $f$ is continuous on $[a, b]$ and $N$ is any value between $f(a)$ and $f(b)$, then there is at least one $c \in (a, b)$ with $f(c) = N$.

In AP problems, the IVT is the go-to justification for "show that $f(x) = 0$ has a solution on $[a, b]$" or "show that $f$ takes the value 5 somewhere on $[a, b]$." Always state the continuity hypothesis explicitly when you cite the IVT — graders will not award the point if you don't.

The Squeeze Theorem

If $g(x) \le f(x) \le h(x)$ near $a$ (except possibly at $a$) and $\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L$, then $\lim_{x\to a} f(x) = L$.

Most-used template: showing that $\lim_{x\to 0} x^{2}\sin(1/x) = 0$ by sandwiching between $-x^{2}$ and $x^{2}$.

How this unit shows up on the AP exam

• Multiple choice (no calculator): Algebraic limit evaluations ($0/0$ form, factoring, rationalizing). Continuity diagnostics from a piecewise definition. Limits-at-infinity / horizontal asymptotes.
• Multiple choice (calculator): Estimating limits from a table or graph; verifying a removable discontinuity numerically.
• Free response: A continuity argument citing IVT; setting up a piecewise function so it's continuous (solve for a parameter); using one-sided limits to characterize a vertical asymptote.

Common mistakes to avoid

• Computing $f(a)$ instead of the limit. They are not the same thing — the limit ignores the value at $a$.
• Saying "$\lim = \infty$" means the limit exists. On the AP exam, an infinite "limit" means the limit does not exist in the formal sense. Use $\pm\infty$ to describe the behavior, not to claim existence.
• Skipping the continuity hypothesis when citing IVT. No "continuous on $[a,b]$" → no credit.
• Plugging $\infty$ into rational functions directly. Always compare degrees first.
• Forgetting the absolute-value subtlety. $\lim_{x \to 0} \frac{|x|}{x}$ does not exist (left = $-1$, right = ).

Quick reference card

• Limit exists $\Leftrightarrow$ left limit = right limit = same finite number
• 3-part continuity: $f(a)$ defined; $\lim$ exists; they're equal
• Discontinuity types: removable / jump / infinite
• Indeterminate forms to attack: $0/0$, $\infty/\infty$ (then factor / rationalize / L'Hôpital)
• Special trig: $\lim_{x\to 0} \sin(x)/x = 1$; $\lim_{x\to 0} (1 - \cos x)/x = 0$
• Rational function at $\infty$: compare leading-term degrees
• IVT requires continuity on a closed interval

Sub-topics in this unit

Use the cards below to drill into each granular skill. Start with the conceptual ones (what a limit is, notation, one-sided limits) and move into the algebraic-technique sections (factoring, rationalizing, indeterminate forms) before tackling continuity and limits at infinity. There's also an interactive lesson and entrance quiz at the top of this page that test the whole unit at once.

So $\lim_{x \to 2} \dfrac{x^{2} - 4}{x - 2} = \lim_{x \to 2}(x + 2) = \boxed{4}$.

Note that $f(2)$ itself is undefined (the original function has a removable discontinuity at $x = 2$), but the limit exists.

Right limit: $\lim_{x \to 1^{+}} f(x) = 4 - 1 = 3$.

Since both one-sided limits equal 3, $\lim_{x \to 1} f(x) = 3$.

$f(1) = 5$ (from the middle piece).

The limit exists and $f(1)$ is defined, but $\lim_{x \to 1} f(x) \ne f(1)$. Continuity fails the third condition → $f$ is not continuous at $x = 1$. The discontinuity is removable (redefining $f(1) = 3$ would fix it).

$\dfrac{\sqrt{x + 9} - 3}{x} \cdot \dfrac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} = \dfrac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)} = \dfrac{x}{x(\sqrt{x + 9} + 3)} = \dfrac{1}{\sqrt{x + 9} + 3}$.

Now plug in $x = 0$: $\dfrac{1}{\sqrt{9} + 3} = \boxed{\dfrac{1}{6}}$.

Rewrite: $\dfrac{\sin(5x)}{3x} = \dfrac{\sin(5x)}{5x} \cdot \dfrac{5}{3}$.

As $x \to 0$, $5x \to 0$, so $\dfrac{\sin(5x)}{5x} \to 1$.

$\lim = 1 \cdot \dfrac{5}{3} = \boxed{\dfrac{5}{3}}$.

Set equal: $2a + 3 = 3 \Rightarrow 2a = 0 \Rightarrow \boxed{a = 0}$.

Step 2. Compute the endpoints:

• $f(0) = 0^{3} + 0 - 1 = -1$
• $f(1) = 1 + 1 - 1 = 1$

Step 3. $f(0) = -1 < 0 < 1 = f(1)$. So $0$ lies between $f(0)$ and $f(1)$.

Step 4. By the IVT, there exists $c \in (0, 1)$ with $f(c) = 0$. ∎

AP grading note: explicitly stating "continuous on $[0,1]$" is required for full credit.

Divide top and bottom by $x$ (with $x > 0$ so $\sqrt{x^{2}} = x$):

$\dfrac{4x}{\sqrt{x^{2} + 4x} + x} = \dfrac{4}{\sqrt{1 + 4/x} + 1}$.

As $x \to \infty$, $4/x \to 0$:

$\lim = \dfrac{4}{\sqrt{1} + 1} = \boxed{2}$.

Step 2 — Multiply by $x^{2}$ (positive for $x \neq 0$): $-x^{2} \le x^{2}\sin(1/x) \le x^{2}$.

Step 3 — Take limits of the outer functions: $\lim_{x \to 0}(-x^{2}) = 0$ and $\lim_{x \to 0}(x^{2}) = 0$.

Step 4 — Apply the Squeeze Theorem: $0 \le \lim_{x \to 0} x^{2}\sin(1/x) \le 0$, so the limit equals $\boxed{0}$.

Note that $\sin(1/x)$ alone has no limit at 0 (it oscillates wildly), but multiplying by $x^{2}$ damps the oscillation.