Your FREE AP Calculus BC Flashcards 2026 – 200+ Cards

The value a function approaches as its input approaches a number; it can exist where the function is undefined.

The left-hand limit and right-hand limit are equal.

The limit at c exists, $f(c)$ is defined, and $\lim_{x\to c} f(x) = f(c)$.

A hole in the graph: the limit exists but does not equal the function value (or the value is missing).

The left and right limits exist but are not equal.

The function approaches ±∞ near the point (a vertical asymptote).

If f is continuous on $[a,b]$, it takes every value between $f(a)$ and $f(b)$.

If $g(x) \le f(x) \le h(x)$ and $g, h$ share a limit L at c, then $f \to L$ too.

$1$

$0$

Compare degrees: same → ratio of leading coefficients; bigger top → ±∞; bigger bottom → 0.

Resolve by factoring, conjugates, or L'Hospital's Rule.

Where the denominator is 0 (and the numerator isn't), giving an infinite limit.

The value the function approaches as x → ±∞ (an end-behavior limit).

Polynomials are continuous and differentiable everywhere.

BC contains all AB content, so AB-content questions yield a separate AB subscore.

$\cos x$

$-\sin x$

$e^x$

$\dfrac{1}{x}$

$\dfrac{d}{dx} x^n = n x^{n-1}$

$f'(a) = \lim_{h\to 0} \dfrac{f(a+h) - f(a)}{h}$

$f'(a) = \lim_{x\to a} \dfrac{f(x) - f(a)}{x - a}$

Instantaneous rate of change; the slope of the tangent line.

Continuity. (But continuity does NOT imply differentiability.)

At corners, cusps, vertical tangents, and discontinuities.

$(fg)' = f'g + fg'$

$\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$

$\sec^2 x$

$\sec x \tan x$

$-\csc^2 x$

$-\csc x \cot x$

$a^x \ln a$

$0$

$1$

$(c f)' = c f'$

$(f \pm g)' = f' \pm g'$

$y - f(a) = f'(a)(x - a)$.

The negative reciprocal of the tangent slope, $-1/f'(a)$.

$2\pi$.

A tangent touches at one point (instantaneous slope); a secant connects two (average slope).

$\dfrac{f(b) - f(a)}{b - a}$ — the slope of the secant line.

The derivative at a point — the limit of average rates as the interval shrinks.

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\dfrac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

$\dfrac{d}{dx}\sin(5x^3) = \cos(5x^3)\cdot 15x^2$

Differentiate both sides, attach dy/dx to every y-term, then solve for dy/dx.

$\dfrac{dy}{dx} = -\dfrac{x}{y}$

$\dfrac{1}{\sqrt{1 - x^2}}$

$-\dfrac{1}{\sqrt{1 - x^2}}$

$\dfrac{1}{1 + x^2}$

$(f^{-1})'(b) = \dfrac{1}{f'(a)}$ where $f(a) = b$.

Take ln of both sides, then differentiate implicitly — useful for variable exponents like xˣ.

$\dfrac{1}{x}$

$x^x(\ln x + 1)$ (by logarithmic differentiation).

Differentiate dy/dx implicitly again, substituting the first derivative where it appears.

$v(t) = s'(t)$, the derivative of position.

$a(t) = v'(t) = s''(t)$.

$|v(t)|$, the absolute value of velocity.

Velocity and acceleration have the SAME sign.

Velocity and acceleration have OPPOSITE signs.

Differentiate the linking equation with respect to time t, then substitute known values and rates.

For $0/0$ or $\infty/\infty$, $\lim \dfrac{f}{g} = \lim \dfrac{f'}{g'}$.

$L(x) = f(a) + f'(a)(x - a)$.

$dy = f'(x)\, dx$.

When several quantities change with time and are tied by one equation.

Differentiate position → velocity → acceleration; integrate to reverse.

Where f'(x) = 0 or is undefined — a candidate for a relative extremum.

$f'(x) > 0$.

$f'(x) < 0$.

f' changes − to + → relative minimum; + to − → relative maximum.

$f''(x) > 0$.

$f''(x) < 0$.

Where concavity changes — f'' changes sign.

At a critical point: f'' > 0 → minimum; f'' < 0 → maximum; f'' = 0 → inconclusive.

$f'(c) = \dfrac{f(b) - f(a)}{b - a}$ for some c in (a, b).

If $f(a) = f(b)$, there is a c in (a, b) with $f'(c) = 0$.

A continuous function on a closed interval attains an absolute max and min.

Compare f at the critical points and at the two endpoints.

Write the quantity, reduce to one variable, set the derivative to 0, then verify the extremum.

Rate of change of the rate — concavity (and acceleration for motion).

A local (relative) extremum is best nearby; a global (absolute) one is best on the whole domain.

Absolute extrema can occur at the endpoints of a closed interval, not just critical points.

A function whose derivative is the integrand; the indefinite integral, written + C.

Net accumulated change — the signed area under a rate curve.

$\displaystyle\int_a^b f(x)\, dx = F(b) - F(a)$.

$\dfrac{d}{dx}\displaystyle\int_a^x f(t)\, dt = f(x)$.

$\dfrac{d}{dx}\displaystyle\int_a^{g(x)} f(t)\, dt = f(g(x))\cdot g'(x)$.

$\displaystyle\int x^n\, dx = \dfrac{x^{n+1}}{n+1} + C,\ n \ne -1$.

$\ln|x| + C$.

$e^x + C$.

$-\cos x + C$.

$\sin x + C$.

$\tan x + C$.

Reverses the chain rule: substitute u for an inner function so the integral becomes a standard form.

$\displaystyle\int u\, dv = uv - \int v\, du$.

Order for choosing u in by-parts: Logs, Inverse trig, Algebraic, Trig, Exponential.

Split a rational function into simpler fractions that can be integrated individually.

An integral with an infinite limit or unbounded integrand; evaluated as a limit.

Underestimates the integral.

Overestimates the integral.

Averages the left and right sums; accurate for nearly linear data.

Uses the height at the midpoint of each subinterval.

$g(x) = \displaystyle\int_a^x f(t)\, dt$ gives net change of f's antiderivative from a to x.

$\displaystyle\int_a^b f'(x)\, dx = f(b) - f(a)$ — integrating a rate gives total change.

$\dfrac{1}{b - a}\displaystyle\int_a^b f(x)\, dx$.

$\arctan x + C$.

$0$ (by symmetry).

$\displaystyle\int_a^a f(x)\, dx = 0$.

$\displaystyle\int_a^b f = -\displaystyle\int_b^a f$.

Approximates a definite integral by summing areas of rectangles.

$kx + C$.

Converges if the defining limit is a finite number; otherwise it diverges.

There is a c with $f(c)$ equal to the average value of f on [a, b].

The net change in the quantity over the interval (the Net Change Theorem).

An equation involving a derivative; a solution is a function satisfying it.

A grid of short segments showing the slope dy/dx of solutions at many points.

Separate variables (y on one side, x on the other), integrate both sides, apply the initial condition.

General has + C; a particular solution uses an initial condition to find C.

$y_{\text{new}} = y_{\text{old}} + \dfrac{dy}{dx}\cdot \Delta x$.

A repeated tangent-line step; smaller step size Δx gives better accuracy.

$\dfrac{dy}{dt} = ky \Rightarrow y = y_0 e^{kt}$.

$\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{M}\right)$, with carrying capacity M.

At $P = M/2$, half the carrying capacity.

$P \to M$ (the carrying capacity) as $t \to \infty$.

$\dfrac{dT}{dt} = k(T - T_s)$; T approaches the surrounding temperature $T_s$.

Integrate to a general solution, then use the initial condition to find C.

$\displaystyle\int_a^b \sqrt{1 + (f'(x))^2}\, dx$

$\displaystyle\int_a^b (\text{top} - \text{bottom})\, dx$.

$\pi\displaystyle\int_a^b [R(x)]^2\, dx$.

$\pi\displaystyle\int_a^b \big([R(x)]^2 - [r(x)]^2\big)\, dx$.

$2\pi\displaystyle\int_a^b (\text{radius})(\text{height})\, dx$.

$\displaystyle\int_a^b A(x)\, dx$, the integral of the cross-sectional area.

$\dfrac{1}{b - a}\displaystyle\int_a^b f(x)\, dx$.

$\displaystyle\int_a^b \sqrt{1 + (f'(x))^2}\, dx$.

$\displaystyle\int_a^b |v(t)|\, dt$ (integrate speed, not velocity).

$\displaystyle\int_a^b v(t)\, dt$ (integrate velocity).

Integrate the rate over the interval to get the total accumulated quantity.

$\dfrac{1}{2}\displaystyle\int_\alpha^\beta [r(\theta)]^2\, d\theta$

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

$\dfrac{d^2 y}{dx^2} = \dfrac{d/dt(dy/dx)}{dx/dt}$.

$\sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}$.

$\displaystyle\int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2}\, dt$.

Position $(x(t), y(t))$; velocity $(x'(t), y'(t))$; acceleration $(x''(t), y''(t))$.

$x = r\cos\theta,\ y = r\sin\theta$.

$r^2 = x^2 + y^2,\ \tan\theta = y/x$.

$\dfrac{1}{2}\displaystyle\int_\alpha^\beta [r(\theta)]^2\, d\theta$.

$\dfrac{1}{2}\displaystyle\int_\alpha^\beta (r_{\text{out}}^2 - r_{\text{in}}^2)\, d\theta$.

Convert to $x = r\cos\theta,\ y = r\sin\theta$, then use the parametric slope in θ.

A polar curve like $r = 1 + \cos\theta$ — a heart shape symmetric about the x-axis.

A circle of radius 1 centered at (1, 0).

$|v(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}$ — the particle's speed.

Integrate each component of the velocity vector and add the initial position.

Replace θ with −θ (x-axis), π − θ (y-axis), or θ + π (origin) and check if the equation is unchanged.

Eliminate the parameter t by solving one equation for t and substituting.

Occurs when both dx/dt and dy/dt are 0 at the same t.

$\dfrac{a}{1 - r}$ when $|r| < 1$

Converges if $\lim |a_{n+1}/a_n| < 1$; diverges if > 1; inconclusive if = 1.

Its terms approach a single finite limit as n → ∞.

Its sequence of partial sums approaches a finite limit.

$\sum ar^n$ converges to $\dfrac{a}{1 - r}$ iff $|r| < 1$.

$\sum \dfrac{1}{n^p}$ converges iff $p > 1$.

$\sum \dfrac{1}{n}$ — diverges (the p = 1 case), even though terms → 0.

If $a_n \not\to 0$, the series diverges. (Can't prove convergence.)

$L = \lim |a_{n+1}/a_n|$: L < 1 converges, L > 1 diverges, L = 1 inconclusive.

$L = \lim \sqrt[n]{|a_n|}$: L < 1 converges, L > 1 diverges, L = 1 inconclusive.

If f is positive, decreasing, continuous, $\sum a_n$ and $\int f$ converge or diverge together.

Compare a positive series term-by-term to a known convergent or divergent series.

If $\lim a_n/b_n$ is a positive finite number, both series share the same fate.

If terms alternate sign, decrease in size, and approach 0, the series converges.

$\sum |a_n|$ converges; this implies $\sum a_n$ converges.

$\sum a_n$ converges but $\sum |a_n|$ diverges (e.g. alternating harmonic).

The error is at most the absolute value of the first omitted term.

$\sum c_n (x - a)^n$, centered at a.

Distance from the center within which a power series converges; found via the Ratio Test.

The set of x for which the power series converges — test endpoints separately.

$\sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!}(x - a)^n$.

A Taylor series centered at a = 0.

$\sum_{n=0}^{\infty} \dfrac{x^n}{n!} = 1 + x + \dfrac{x^2}{2!} + \cdots$

$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$

$1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$

$\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots,\ |x| < 1$

A finite truncation of a Taylor series that approximates f near the center.

Bounds how far a Taylor polynomial can be from the true value of f.

A series whose partial sums collapse because consecutive terms cancel.

Geometric with a = 4/10, r = 1/10 → $\dfrac{4/10}{1 - 1/10} = \dfrac{4}{9}$.

Differentiate, integrate, or substitute into a known Maclaurin series to get a new one.

Check nth-Term first, recognize geometric/p-series, then Ratio, Comparison, Alternating.

The Ratio Test gives L = 1 at the endpoints, so it is inconclusive there.

$x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots,\ |x| \le 1$.

$x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots,\ -1 < x \le 1$.

The sum of the first n terms of a series; its limit (if finite) is the series sum.

A sequence that is bounded and monotonic converges.

$a_n = a_1 r^{n-1}$.

Terms going to 0 is necessary but not sufficient — e.g. the harmonic series.

Compare a messy positive series to 1/nᵖ to decide convergence quickly.

Factorials and nth powers make the Ratio Test the natural first choice.

The center a is where you take the derivatives; choose it near the x-values you'll evaluate.

Radius 1; converges for $|x| < 1$.

$\sum (-1)^{n+1}/n$ converges (conditionally) to $\ln 2$.