Limits & Continuity
Memory Anchor
0/0 or ∞/∞ → differentiate top & bottom separately
"When stuck at 0/0, derive your way out."
"When stuck at 0/0, derive your way out."
Evaluate: \[\lim_{x \to 0} \frac{\sin(3x) - 3x}{x^3}\]
Step-by-step Explanation
This is a \(\frac{0}{0}\) form. Apply L'Hôpital's Rule three times:
1st derivative: \(\dfrac{3\cos(3x)-3}{3x^2}\) → still \(\frac{0}{0}\)
2nd: \(\dfrac{-9\sin(3x)}{6x}\) → still \(\frac{0}{0}\)
3rd: \(\dfrac{-27\cos(3x)}{6}\)
As \(x\to 0\): \(\dfrac{-27\cdot1}{6} = \boxed{-\dfrac{9}{2}}\)
1st derivative: \(\dfrac{3\cos(3x)-3}{3x^2}\) → still \(\frac{0}{0}\)
2nd: \(\dfrac{-9\sin(3x)}{6x}\) → still \(\frac{0}{0}\)
3rd: \(\dfrac{-27\cos(3x)}{6}\)
As \(x\to 0\): \(\dfrac{-27\cdot1}{6} = \boxed{-\dfrac{9}{2}}\)
Memory Anchor
Trap it: L ≤ f(x) ≤ U → if L = U at a point, f must go there too
"Squeeze the middle — it has nowhere to go."
"Squeeze the middle — it has nowhere to go."
Given that \(-x^2 \leq f(x) \leq x^2\) for all \(x\), find \(\displaystyle\lim_{x \to 0} f(x)\).
Explanation
Since \(\lim_{x\to 0}(-x^2) = 0\) and \(\lim_{x\to 0}(x^2) = 0\), by the Squeeze Theorem, \(\lim_{x\to 0} f(x) = \boxed{0}\).
Differentiation
Memory Anchor
Every y-term gets a dy/dx rider via chain rule
"d/dx of y = (dy/dx). Always attach the trailer."
"d/dx of y = (dy/dx). Always attach the trailer."
If \(x^3 + y^3 = 6xy\), find \(\dfrac{dy}{dx}\).
Explanation
Differentiate both sides: \(3x^2 + 3y^2\tfrac{dy}{dx} = 6y + 6x\tfrac{dy}{dx}\)
Collect \(\tfrac{dy}{dx}\): \(\tfrac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2\)
\(\dfrac{dy}{dx} = \dfrac{6y-3x^2}{3y^2-6x} = \boxed{\dfrac{2y-x^2}{y^2-2x}}\)
Collect \(\tfrac{dy}{dx}\): \(\tfrac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2\)
\(\dfrac{dy}{dx} = \dfrac{6y-3x^2}{3y^2-6x} = \boxed{\dfrac{2y-x^2}{y^2-2x}}\)
Memory Anchor
Outer′(inner) × inner′ — peel the onion layer by layer
Let \(f(x) = \sin^2(\cos(x^2))\). Find \(f'(x)\).
Explanation — Three Layers
Layer 1: \((\cdot)^2\) → brings down 2, keeps inside
Layer 2: \(\sin(\cdot)\) → becomes \(\cos(\cdot)\)
Layer 3: \(\cos(x^2)\) → becomes \(-\sin(x^2)\cdot 2x\)
Combined: \(f'(x) = 2\sin(\cos x^2)\cdot\cos(\cos x^2)\cdot(-\sin x^2)\cdot 2x\)
\(= \boxed{-4x\sin(\cos x^2)\cos(\cos x^2)\sin(x^2)}\)
Layer 2: \(\sin(\cdot)\) → becomes \(\cos(\cdot)\)
Layer 3: \(\cos(x^2)\) → becomes \(-\sin(x^2)\cdot 2x\)
Combined: \(f'(x) = 2\sin(\cos x^2)\cdot\cos(\cos x^2)\cdot(-\sin x^2)\cdot 2x\)
\(= \boxed{-4x\sin(\cos x^2)\cos(\cos x^2)\sin(x^2)}\)
Memory Anchor
Write the equation → differentiate both sides with respect to t → plug in known values LAST
A spherical balloon is being inflated so that its volume increases at \(10 \text{ cm}^3/\text{s}\). How fast is the radius increasing when the radius is \(5\) cm?
\(\left[V = \tfrac{4}{3}\pi r^3\right]\)
\(\left[V = \tfrac{4}{3}\pi r^3\right]\)
Explanation
\(\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}\)
Plug in: \(10 = 4\pi(25)\dfrac{dr}{dt}\)
\(\dfrac{dr}{dt} = \dfrac{10}{100\pi} = \boxed{\dfrac{1}{10\pi}}\) cm/s
Plug in: \(10 = 4\pi(25)\dfrac{dr}{dt}\)
\(\dfrac{dr}{dt} = \dfrac{10}{100\pi} = \boxed{\dfrac{1}{10\pi}}\) cm/s
Integration Techniques
Memory Anchor
LIATE — pick u in this order: Logs, Inverse trig, Algebraic, Trig, Exponential
"∫u dv = uv − ∫v du"
"∫u dv = uv − ∫v du"
Evaluate \(\displaystyle\int x e^x \, dx\).
Explanation
Let \(u = x\), \(dv = e^x dx\) → \(du = dx\), \(v = e^x\)
\(\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C = \boxed{e^x(x-1)+C}\)
\(\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C = \boxed{e^x(x-1)+C}\)
Memory Anchor
Factor denominator → decompose → solve for A, B by substituting roots
\[\int \frac{1}{x^2 - 1} \, dx = ?\]
Explanation
\(\dfrac{1}{x^2-1} = \dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1}+\dfrac{B}{x+1}\)
Solving: \(A = \frac{1}{2}, B = -\frac{1}{2}\)
\(\int = \tfrac{1}{2}\ln|x-1| - \tfrac{1}{2}\ln|x+1| + C = \boxed{\tfrac{1}{2}\ln\left|\tfrac{x-1}{x+1}\right|+C}\)
Solving: \(A = \frac{1}{2}, B = -\frac{1}{2}\)
\(\int = \tfrac{1}{2}\ln|x-1| - \tfrac{1}{2}\ln|x+1| + C = \boxed{\tfrac{1}{2}\ln\left|\tfrac{x-1}{x+1}\right|+C}\)
Memory Anchor
√(a²−x²) → x=a sinθ · √(a²+x²) → x=a tanθ · √(x²−a²) → x=a secθ
\[\int \frac{dx}{\sqrt{9 - x^2}}\]
Explanation
Let \(x = 3\sin\theta\), \(dx = 3\cos\theta\,d\theta\), \(\sqrt{9-x^2}=3\cos\theta\)
\(\int\dfrac{3\cos\theta\,d\theta}{3\cos\theta} = \int d\theta = \theta + C = \boxed{\arcsin\!\left(\tfrac{x}{3}\right)+C}\)
\(\int\dfrac{3\cos\theta\,d\theta}{3\cos\theta} = \int d\theta = \theta + C = \boxed{\arcsin\!\left(\tfrac{x}{3}\right)+C}\)
Infinite Series
Memory Anchor
L = lim |a(n+1)/a(n)| → L<1: converges · L>1: diverges · L=1: inconclusive
Does \(\displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^n}\) converge or diverge? (Use the Ratio Test)
Explanation
\(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^n}{n!} = \left(\dfrac{n}{n+1}\right)^n = \dfrac{1}{\left(1+\frac{1}{n}\right)^n} \to \dfrac{1}{e}\)
Since \(L = \frac{1}{e} < 1\), the series converges by the Ratio Test.
Since \(L = \frac{1}{e} < 1\), the series converges by the Ratio Test.
Memory Anchor
Know these cold: e^x = Σxⁿ/n!, sin x = Σ(−1)ⁿx^(2n+1)/(2n+1)!, cos x = Σ(−1)ⁿx^(2n)/(2n)!
The Maclaurin series for \(\sin x\) is used to find the series for \(\sin(x^2)\). The first three nonzero terms are:
Explanation
Substitute \(x^2\) for \(x\) in \(\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots\)
\(\sin(x^2) = x^2 - \dfrac{(x^2)^3}{6} + \dfrac{(x^2)^5}{120} - \cdots = \boxed{x^2 - \dfrac{x^6}{6} + \dfrac{x^{10}}{120} - \cdots}\)
\(\sin(x^2) = x^2 - \dfrac{(x^2)^3}{6} + \dfrac{(x^2)^5}{120} - \cdots = \boxed{x^2 - \dfrac{x^6}{6} + \dfrac{x^{10}}{120} - \cdots}\)
Memory Anchor
Use Ratio Test on power series → solve |x − c| < R → R is radius. Always check endpoints separately!
Find the radius of convergence of \(\displaystyle\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}\).
Explanation
Ratio Test: \(\left|\dfrac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\dfrac{n\cdot 3^n}{(x-2)^n}\right| = \dfrac{n}{n+1}\cdot\dfrac{|x-2|}{3} \to \dfrac{|x-2|}{3}\)
For convergence: \(\dfrac{|x-2|}{3} < 1 \Rightarrow |x-2| < 3\), so \(\boxed{R = 3}\).
For convergence: \(\dfrac{|x-2|}{3} < 1 \Rightarrow |x-2| < 3\), so \(\boxed{R = 3}\).
Memory Anchor
AST: terms must be (1) decreasing in absolute value AND (2) approaching zero → converges
Which series converges by the Alternating Series Test?
Explanation
A: terms → \(\frac{n}{n+1}\to 1 \neq 0\) ✗
B: \(\frac{1}{\sqrt{n}}\) is decreasing and \(\to 0\) ✓ → Converges
C: terms don't approach 0 ✗
D: \(n!/n^2 \to \infty\) ✗
B: \(\frac{1}{\sqrt{n}}\) is decreasing and \(\to 0\) ✓ → Converges
C: terms don't approach 0 ✗
D: \(n!/n^2 \to \infty\) ✗
Differential Equations
Memory Anchor
Move all y's left, all x's right → integrate both sides → solve for y
Solve: \(\dfrac{dy}{dx} = xy\), with \(y(0) = 3\).
Explanation
\(\dfrac{dy}{y} = x\,dx\)
\(\ln|y| = \dfrac{x^2}{2} + C\)
\(y = Ae^{x^2/2}\)
\(y(0)=3 \Rightarrow A=3\)
\(\boxed{y = 3e^{x^2/2}}\)
\(\ln|y| = \dfrac{x^2}{2} + C\)
\(y = Ae^{x^2/2}\)
\(y(0)=3 \Rightarrow A=3\)
\(\boxed{y = 3e^{x^2/2}}\)
Memory Anchor
dP/dt = kP(1 − P/M) → fastest growth at P = M/2 (the inflection point)
A population satisfies \(\dfrac{dP}{dt} = 0.2P\!\left(1 - \dfrac{P}{500}\right)\). The population grows fastest when \(P =\)
Explanation
\(\dfrac{dP}{dt}\) is maximized when \(\dfrac{d}{dP}\!\left[P(1-P/M)\right] = 0\)
→ \(1 - 2P/M = 0 \Rightarrow P = M/2 = \boxed{250}\)
The inflection point of logistic growth is always at half the carrying capacity.
→ \(1 - 2P/M = 0 \Rightarrow P = M/2 = \boxed{250}\)
The inflection point of logistic growth is always at half the carrying capacity.
Parametric & Polar
Memory Anchor
dy/dx = (dy/dt) ÷ (dx/dt) · Never forget to divide!
A curve is defined parametrically by \(x = t^2 + 1\), \(y = t^3 - 3t\). Find \(\dfrac{dy}{dx}\) when \(t = 2\).
Explanation
\(\dfrac{dx}{dt} = 2t\), \(\dfrac{dy}{dt} = 3t^2 - 3\)
\(\dfrac{dy}{dx} = \dfrac{3t^2-3}{2t}\)
At \(t=2\): \(\dfrac{3(4)-3}{2(2)} = \dfrac{9}{4}\)
\(\dfrac{dy}{dx} = \dfrac{3t^2-3}{2t}\)
At \(t=2\): \(\dfrac{3(4)-3}{2(2)} = \dfrac{9}{4}\)
Memory Anchor
Polar Area = ½∫r² dθ · "Half the square of r, integrate over angle"
Find the area enclosed by \(r = 2\cos\theta\).
Explanation
\(A = \tfrac{1}{2}\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2\,d\theta = \tfrac{1}{2}\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta\)
\(= 2\int_{-\pi/2}^{\pi/2}\dfrac{1+\cos 2\theta}{2}\,d\theta = \int_{-\pi/2}^{\pi/2}(1+\cos 2\theta)\,d\theta\)
\(= [\theta + \tfrac{\sin 2\theta}{2}]_{-\pi/2}^{\pi/2} = \pi\)
The curve \(r = 2\cos\theta\) is a circle of radius 1, area \(= \pi(1)^2 = \boxed{\pi}\).
\(= 2\int_{-\pi/2}^{\pi/2}\dfrac{1+\cos 2\theta}{2}\,d\theta = \int_{-\pi/2}^{\pi/2}(1+\cos 2\theta)\,d\theta\)
\(= [\theta + \tfrac{\sin 2\theta}{2}]_{-\pi/2}^{\pi/2} = \pi\)
The curve \(r = 2\cos\theta\) is a circle of radius 1, area \(= \pi(1)^2 = \boxed{\pi}\).
FTC & Applications
Memory Anchor
d/dx ∫ₐˣ f(t)dt = f(x) · If upper limit is g(x), multiply by g′(x) (chain rule!)
If \(g(x) = \displaystyle\int_1^{x^3} \sqrt{1+t^2}\,dt\), find \(g'(x)\).
Explanation
By FTC Part 1 + Chain Rule:
\(g'(x) = \sqrt{1+(x^3)^2}\cdot\dfrac{d}{dx}(x^3) = \sqrt{1+x^6}\cdot 3x^2 = \boxed{3x^2\sqrt{1+x^6}}\)
\(g'(x) = \sqrt{1+(x^3)^2}\cdot\dfrac{d}{dx}(x^3) = \sqrt{1+x^6}\cdot 3x^2 = \boxed{3x^2\sqrt{1+x^6}}\)
Memory Anchor
Arc Length = ∫√(1+(dy/dx)²)dx · "Square root of 1 plus slope squared, integrated"
Set up (do not evaluate) the arc length of \(y = x^{3/2}\) from \(x=0\) to \(x=4\).
Explanation
\(y' = \dfrac{3}{2}x^{1/2}\), so \((y')^2 = \dfrac{9}{4}x\)
\(L = \int_0^4\sqrt{1+\dfrac{9x}{4}}\,dx\)
\(L = \int_0^4\sqrt{1+\dfrac{9x}{4}}\,dx\)
Memory Anchor
Disk: π∫[f(x)]²dx · Washer: π∫([R(x)]²−[r(x)]²)dx · Shell: 2π∫x·f(x)dx
The region bounded by \(y = \sqrt{x}\), \(y=0\), and \(x=4\) is revolved about the \(x\)-axis. Volume \(=\)
Explanation
Disk method: \(V = \pi\int_0^4(\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\dfrac{x^2}{2}\right]_0^4 = \pi\cdot 8 = \boxed{8\pi}\)
Memory Anchor
Replace ∞ with limit variable b, integrate, then take lim as b→∞. p-series: ∫1/xᵖ converges iff p>1
Evaluate: \(\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx\)
Explanation
\(\int_1^{\infty}x^{-2}dx = \lim_{b\to\infty}\left[-\dfrac{1}{x}\right]_1^b = \lim_{b\to\infty}\left(-\dfrac{1}{b}+1\right) = 0 + 1 = \boxed{1}\)
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