Math Core

BC builds everything on limits: improper integrals are limits, series are limits, and every derivative you take is secretly a limit. This review packs the AB limits toolkit into one page, with the harder problem types BC exams like to reuse.

What a limit says

$\displaystyle\lim_{x \to a} f(x) = L$ means that $f(x)$ can be made as close to $L$ as you like by taking $x$ close enough to $a$ (but not equal to $a$ ). The value $f(a)$ plays no role: it can be different from $L$ or undefined.

The two-sided limit exists exactly when both one-sided limits exist and agree:

\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L \text{ and } \lim_{x \to a^+} f(x) = L.

A limit does not exist (DNE) when the one-sided limits differ, when the function grows without bound (we still write $\lim = \infty$ to describe how it fails), or when it oscillates, as $\sin(1/x)$ does near $0$ .

Evaluating limits algebraically

Always start by substituting. If you get a real number, you are done (for continuous functions). If you get $\dfrac{k}{0}$ with $k \ne 0$ , the limit is infinite or DNE, so check the signs on each side. If you get $\dfrac{0}{0}$ , the limit is indeterminate: more work is needed.

The standard ways to clear a $\dfrac{0}{0}$ :

Situation	Move
Polynomials or rational functions	Factor and cancel the common factor
Square roots	Multiply by the conjugate
Complex fractions	Combine into a single fraction
Trig functions	Rewrite in terms of the two special trig limits
Anything differentiable	L'Hôpital's rule (reviewed in the last lesson of this unit)

Special limits to know cold

\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0, \qquad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}.

More generally, $\displaystyle\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$ for any nonzero constant $k$ .

Worked example: Conjugates and trig limits

Evaluate (a) $\displaystyle\lim_{x \to 5} \frac{\sqrt{x + 4} - 3}{x - 5}$ and (b) $\displaystyle\lim_{x \to 0} \frac{\tan(3x)}{\sin(7x)}$ .

Solution. (a) Substitution gives $\dfrac{0}{0}$ . Multiply by the conjugate:

\frac{\sqrt{x+4} - 3}{x - 5} \cdot \frac{\sqrt{x+4} + 3}{\sqrt{x+4} + 3} = \frac{x - 5}{(x - 5)(\sqrt{x+4} + 3)} = \frac{1}{\sqrt{x+4} + 3} \to \frac{1}{6}.

(b) Write $\tan(3x) = \dfrac{\sin(3x)}{\cos(3x)}$ and build the special limit twice:

\frac{\tan 3x}{\sin 7x} = \frac{\sin 3x}{3x} \cdot \frac{7x}{\sin 7x} \cdot \frac{3}{7} \cdot \frac{1}{\cos 3x} \to 1 \cdot 1 \cdot \frac{3}{7} \cdot 1 = \frac{3}{7}.

The squeeze theorem

If $g(x) \le f(x) \le h(x)$ near $a$ and $g$ and $h$ both approach $L$ , then $f(x) \to L$ too. The classic use is a bounded factor times something going to zero. For example, $-x^2 \le x^2 \cos(1/x) \le x^2$ , and both bounds go to $0$ , so $\displaystyle\lim_{x\to 0} x^2\cos(1/x) = 0$ .

Limits at infinity and asymptotes

For a rational function $\dfrac{p(x)}{q(x)}$ as $x \to \pm\infty$ , compare degrees:

degree of top smaller: the limit is $0$ ;
degrees equal: the limit is the ratio of leading coefficients;
degree of top larger: the limit is $\pm\infty$ (no horizontal asymptote).

A finite limit at $\pm\infty$ gives a horizontal asymptote. An infinite one-sided limit at $x = a$ gives a vertical asymptote. For rational functions, cancel common factors first: a factor that cancels gives a hole, not an asymptote.

Growth rates settle many limits at infinity at a glance: as $x \to \infty$ ,

\ln x \ll x^p \ll b^x \ll x! \quad (p > 0,\ b > 1),

where $\ll$ means "is eventually negligible compared to." So $\dfrac{x^{10}}{e^x} \to 0$ and $\dfrac{\ln x}{\sqrt{x}} \to 0$ . You will use this ordering constantly in the series unit.

Common mistake

When $x \to -\infty$ , $\sqrt{x^2} = |x| = -x$ , not $x$ . Forgetting this sign flips the answer on every limit that divides a square root by $x$ .

Worked example: Radicals at negative infinity

Evaluate $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{9x^2 + 2x}}{4x - 1}$ .

Solution. Divide top and bottom by $|x| = -x$ (since $x$ is negative):

\frac{\sqrt{9x^2 + 2x}}{4x - 1} = \frac{\sqrt{9 + 2/x}}{\dfrac{4x - 1}{-x}} = \frac{\sqrt{9 + 2/x}}{-4 + 1/x} \to \frac{3}{-4} = -\frac{3}{4}.

Continuity

Definition

Continuity at a point

$f$ is continuous at $x = a$ when all three hold: $f(a)$ is defined, $\displaystyle\lim_{x\to a} f(x)$ exists, and $\displaystyle\lim_{x\to a} f(x) = f(a)$ .

The three ways to fail give the three types of discontinuity:

Removable (a hole): the limit exists but $f(a)$ is missing or wrong.
Jump: the one-sided limits exist but differ.
Infinite: at least one one-sided limit is infinite (a vertical asymptote).

Polynomials, rational functions, roots, exponentials, logs and trig functions are continuous on their domains, and so are sums, products, quotients and compositions of them. The only places to check are domain gaps and the seams of piecewise definitions.

A removable discontinuity: the limit at x = 2 is 4, but f(2) was redefined as 1.Open in grapher →

Worked example: Two unknowns at a seam

Find $a$ and $b$ so that $f$ is continuous everywhere:

f(x) = \begin{cases} ax + 3, & x < 1 \\ x^2 + b, & 1 \le x \le 3 \\ 2ax - b, & x > 3. \end{cases}

Solution. Match the pieces at each seam. At $x = 1$ : $a + 3 = 1 + b$ . At $x = 3$ : $9 + b = 6a - b$ , so $6a - 2b = 9$ . From the first equation $b = a + 2$ ; substituting, $6a - 2a - 4 = 9$ , so $a = \dfrac{13}{4}$ and $b = \dfrac{21}{4}$ .

The Intermediate Value Theorem

Intermediate Value Theorem (IVT)

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

On free-response questions, always name the hypothesis: "Because $f$ is continuous (it is differentiable) on $[a, b]$ and $f(a) < N < f(b)$ , the IVT guarantees..." The IVT only guarantees existence; it never tells you where $c$ is or how many there are beyond the minimum count.

Tip

To count guaranteed zeros from a table, count the sign changes between consecutive listed values. Each sign change of a continuous function forces at least one zero in that interval.