lim[x→c] f(x) = L  (two-sided limit)
Squeeze Theorem: g(x) ≤ f(x) ≤ h(x), lim g = lim h = L → lim f = L
lim[x→0] sin(x)/x = 1    // must memorize
lim[x→0] (1−cos x)/x = 0
lim[x→±∞] (aₙxⁿ + …)/(bₙxⁿ + …) = aₙ/bₙ  // same degree

f'(x) = lim[h→0] [f(x+h)−f(x)] / h

Power Rule:  d/dx[xⁿ] = nxⁿ⁻¹
Product Rule: d/dx[uv] = u'v + uv'
Quotient Rule: d/dx[u/v] = (u'v − uv') / v²
Chain Rule:  d/dx[f(g(x))] = f'(g(x))·g'(x)

Differentiate both sides w.r.t. x;
d/dx[y²] = 2y · (dy/dx)   // chain rule on y
Then solve algebraically for dy/dx.

MVT: f'(c) = [f(b)−f(a)] / (b−a)  for some c ∈ (a,b)
Conditions: continuous on [a,b], differentiable on (a,b)
Rolle's: f(a)=f(b) → f'(c)=0 for some c

FTC Part 1: d/dx[∫[a,x] f(t)dt] = f(x)
FTC Part 2: ∫[a,b] f(x)dx = F(b)−F(a)

∫xⁿ dx = xⁿ⁺¹/(n+1) + C  (n≠−1)
∫eˣ dx = eˣ + C
∫(1/x) dx = ln|x| + C
∫sin x dx = −cos x + C
∫cos x dx = sin x + C
∫sec²x dx = tan x + C

u-sub: ∫f(g(x))g'(x)dx → ∫f(u)du (set u=g(x))

Parts (BC): ∫u dv = uv − ∫v du
// LIATE priority: Log, Inverse trig, Algebraic, Trig, Exp

Area between curves: ∫[a,b] [f(x)−g(x)] dx (f ≥ g)
Disk method: V = π∫[a,b] [R(x)]² dx
Washer method: V = π∫[a,b] ([R(x)]²−[r(x)]²) dx
Shell method: V = 2π∫[a,b] x·f(x) dx

Separable: dy/dx = f(x)g(y) → ∫dy/g(y) = ∫f(x)dx
Exponential model: dP/dt = kP → P = P₀eᵏᵗ
Logistic: dP/dt = kP(1−P/L),  L = carrying capacity

Geometric series: Σ arⁿ = a/(1−r), |r|<1
p-series: Σ 1/nᵖ converges iff p>1
Ratio Test: L = lim|aₙ₊₁/aₙ|; conv if L<1, div if L>1
AST: alternating series converges if |aₙ| decreasing → 0

Taylor: f(x) = Σ [f⁽ⁿ⁾(a)/n!](x−a)ⁿ

eˣ = 1 + x + x²/2! + x³/3! + …  (all x)
sin x = x − x³/3! + x⁵/5! − …    (all x)
cos x = 1 − x²/2! + x⁴/4! − …    (all x)
1/(1−x) = 1 + x + x² + x³ + …   |x|<1