AP^® Calculus AB & BC

THE BIG PICTURE. Unit 1 introduces the single most important concept in calculus: the limit. Every subsequent idea (derivatives, integrals, series) is built on the limit. The unit weighs 10–12% of AB / 4–7% of BC but its conceptual importance is enormous: students who don't internalize limits as the formal way to talk about "approaches" struggle with everything that follows. The MC and FRQ both directly test (1) computing limits, (2) continuity classifications, and (3) the IVT for proving roots exist.

Notation: $\lim_{x \to c} f(x) = L$ means as $x$ gets arbitrarily close to $c$ (from either side), $f(x)$ gets arbitrarily close to $L$.

One-sided limits:

• $\lim_{x \to c^-} f(x)$: limit from the LEFT (values less than $c$).
• $\lim_{x \to c^+} f(x)$: limit from the RIGHT (values greater than $c$).

The two-sided limit $\lim_{x \to c} f(x)$ EXISTS only when both one-sided limits exist and AGREE.

◆ DIRECT SUBSTITUTION: works whenever $f$ is continuous at $c$.

• Example: $\lim_{x \to 2} (x^2 + 3x) = 4 + 6 = 10$.

◆ ALGEBRAIC SIMPLIFICATION: when direct substitution gives an INDETERMINATE FORM $\frac{0}{0}$:

• FACTOR and CANCEL. $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 6$.
• RATIONALIZE (multiply by conjugate). $\lim_{x \to 0} \frac{\sqrt{x+4}-2}{x} = \lim_{x \to 0} \frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{4}$.
• COMMON DENOMINATORS: when limit involves complex fractions.

◆ SPECIAL LIMITS to memorize:

• $\lim_{x \to 0} \frac{\sin x}{x} = 1$
• $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$
• $\lim_{x \to 0} \frac{\tan x}{x} = 1$
• $\lim_{x \to 0} (1 + x)^{1/x} = e$

Three discontinuity types:

• REMOVABLE (hole): limit exists but $\neq f(c)$, or $f(c)$ is undefined.
• JUMP: one-sided limits exist but differ.
• INFINITE: function approaches $\pm\infty$ at the point (vertical asymptote).

Continuity FACTS: polynomials, $\sin x$, $\cos x$, $e^x$ are continuous everywhere. Rational functions are continuous wherever their denominator is nonzero. $\tan x$, $\sec x$ have infinite discontinuities at the zeros of $\cos x$.

For RATIONAL functions $\frac{p(x)}{q(x)}$, compare degrees:

• Numerator degree < denominator degree: $\lim = 0$. Horizontal asymptote at $y = 0$.
• Numerator degree = denominator degree: $\lim = \frac{\text{leading coeff of } p}{\text{leading coeff of } q}$. Horizontal asymptote at that ratio.
• Numerator degree > denominator degree: $\lim = \pm\infty$. No horizontal asymptote. (May have a SLANT (oblique) asymptote if numerator is exactly one degree higher.)

◆ INTERMEDIATE VALUE THEOREM (IVT).

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

Most common AP use: prove a function has a ROOT in $[a, b]$ by showing $f$ is continuous and $f(a)$ and $f(b)$ have opposite signs (taking $N = 0$).

Required setup language on FRQ: "$f$ is continuous on $[a, b]$": state this explicitly before invoking IVT.

◆ SQUEEZE (SANDWICH) THEOREM.

If $g(x) \leq f(x) \leq h(x)$ near $x = c$, and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$.

Classic application: $\lim_{x \to 0} x^2 \sin\!\left(\tfrac{1}{x}\right) = 0$ (because $-x^2 \leq x^2 \sin(1/x) \leq x^2$).

EXAM CONNECTIONS. Limits appear directly on every exam through MC questions on computation, classification of discontinuities, end behavior, and IVT setups. They appear INDIRECTLY on every derivative and integral question: the formal definitions of both are LIMITS. Master Unit 1 and the rest of the course rests on solid foundation.