Section 01
Limits & Continuity
P-01
FOUNDATIONAL
Limits · L'Hôpital's Rule
Evaluating Indeterminate Limits
⚡ Quick Keys
0/0 → L'Hôpital
∞/∞ → L'Hôpital
factor first
conjugate trick
Let \(f(x) = \dfrac{x^2 - 9}{x - 3}\). A student claims \(f\) is continuous at \(x = 3\) because the limit exists.
Consider the function \(g(x) = \dfrac{\sin(2x)}{x}\).
Consider the function \(g(x) = \dfrac{\sin(2x)}{x}\).
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(a)Find \(\displaystyle\lim_{x \to 3} f(x)\). Show your reasoning.ExplanationFactor: \(x^2 - 9 = (x-3)(x+3)\), so \(\dfrac{(x-3)(x+3)}{x-3} = x+3\) for \(x \ne 3\).
Therefore \(\lim_{x\to3}f(x) = 3 + 3 = \mathbf{6}\). Note: \(f(3)\) is undefined, so \(f\) is not continuous at \(x=3\) — the student is wrong. A limit existing does not guarantee continuity; the function value must also equal the limit. -
(b)Find \(\displaystyle\lim_{x \to 0} g(x) = \lim_{x \to 0}\frac{\sin(2x)}{x}\).ExplanationUse the standard limit \(\lim_{u\to0}\frac{\sin u}{u}=1\).
\(\frac{\sin(2x)}{x} = 2\cdot\frac{\sin(2x)}{2x} \to 2\cdot 1 = \mathbf{2}\).
Or apply L'Hôpital: \(\frac{2\cos(2x)}{1}\Big|_{x=0} = 2\). -
(c)Is \(f(x) = \dfrac{x^2-9}{x-3}\) continuous at \(x = 3\)? Justify completely using the three-part definition.ExplanationThree conditions for continuity at \(x = c\):
1. \(f(c)\) is defined 2. \(\lim_{x\to c}f(x)\) exists 3. \(\lim_{x\to c}f(x) = f(c)\)
Here \(f(3)\) is undefined (denominator = 0, no cancellation saves the value), so condition 1 fails. The function has a removable discontinuity at \(x = 3\).
🔀 Variant
Try: \(h(x) = \dfrac{e^{2x}-1}{\sin x}\) as \(x\to 0\). Use L'Hôpital and verify against the standard limit \(\lim_{x\to0}\frac{e^{kx}-1}{x}=k\).
P-02
INTERMEDIATE
Limits · Squeeze Theorem
Bounding Oscillating Functions
⚡ Quick Keys
sandwich it
−1 ≤ sin ≤ 1
multiply bounds
same limit → squeeze
Consider \(f(x) = x^2 \sin\!\left(\dfrac{1}{x}\right)\) for \(x \ne 0\).
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(a)State an appropriate inequality that allows the Squeeze Theorem to apply to find \(\displaystyle\lim_{x\to 0}f(x)\).ExplanationSince \(-1 \le \sin(1/x) \le 1\) for all \(x\ne0\), multiply through by \(x^2 \ge 0\):
\(-x^2 \le x^2\sin(1/x) \le x^2\).
Both bounds approach 0 as \(x\to0\), so by Squeeze Theorem, \(\lim_{x\to0}x^2\sin(1/x) = 0\). -
(b)Define \(f(0) = 0\). Is \(f\) differentiable at \(x = 0\)? Evaluate \(\lim_{h\to0}\dfrac{f(h)-f(0)}{h}\).Explanation\(\dfrac{f(h)-0}{h} = \dfrac{h^2\sin(1/h)}{h} = h\sin(1/h)\).
By Squeeze: \(-|h| \le h\sin(1/h) \le |h| \to 0\).
So \(f'(0) = 0\). Note: contrast with \(x\sin(1/x)\) at 0, which is continuous but not differentiable.
🔀 Variant
Consider \(g(x) = x\sin(1/x)\), \(g(0)=0\). Show \(g\) is continuous at 0 using Squeeze, then show \(g'(0)\) does not exist.
P-03
EXAM LEVEL
Limits · Improper & Asymptotic Behavior
Limits at Infinity and Horizontal Asymptotes
⚡ Quick Keys
divide by highest power
e^x beats poly
ln x vs x^n
HA: lim at ±∞
The function \(R(x) = \dfrac{3x^3 - 2x + 1}{5x^3 + x^2 - 7}\) models a ratio in a physics experiment.
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(a)Determine \(\displaystyle\lim_{x\to\infty} R(x)\) and \(\displaystyle\lim_{x\to-\infty}R(x)\). Identify any horizontal asymptotes.ExplanationDivide numerator and denominator by \(x^3\):
\(\lim_{x\to\pm\infty}\dfrac{3-2/x^2+1/x^3}{5+1/x-7/x^3} = \dfrac{3}{5}\).
Since both limits are equal, there is exactly one horizontal asymptote: \(y = \dfrac{3}{5}\). -
(b)Evaluate \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{x^{0.1}}\). Which grows faster?ExplanationL'Hôpital (\(\infty/\infty\)): \(\dfrac{(1/x)}{0.1\,x^{-0.9}} = \dfrac{1}{0.1\,x^{0.1}} = \dfrac{10}{x^{0.1}} \to 0\).
Key hierarchy: \(\ln x \ll x^p\) for any \(p > 0\). Any positive power of \(x\) eventually dominates \(\ln x\).
🔀 Variant
Analyze \(\lim_{x\to\infty}\dfrac{x^{100}}{e^x}\). Apply L'Hôpital repeatedly (or use the dominance rule: \(e^x\) beats every polynomial). What does this tell you about exponential vs. polynomial growth?
Section 02
Differentiation
P-04
FOUNDATIONAL
Derivatives · Chain Rule
Composite & Implicit Differentiation
⚡ Quick Keys
outside·inside'
implicit: d/dx of y² = 2y·y'
product rule
dy/dx = −Fx/Fy
Let \(F(x) = \sin^3(2x^2 + 1)\). Also consider the curve \(x^2 y + y^3 = 10\).
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(a)Find \(F'(x)\).ExplanationChain rule twice. Let \(u = 2x^2+1\), \(v = \sin u\), \(F = v^3\).
\(F' = 3v^2 \cdot v' = 3\sin^2(u)\cdot\cos(u)\cdot u' = 3\sin^2(2x^2+1)\cos(2x^2+1)\cdot 4x\)
\(= 12x\sin^2(2x^2+1)\cos(2x^2+1)\). -
(b)Use implicit differentiation to find \(\dfrac{dy}{dx}\) for \(x^2y + y^3 = 10\).ExplanationDifferentiate both sides: \(\frac{d}{dx}[x^2y] + \frac{d}{dx}[y^3] = 0\).
Product rule on first term: \(2xy + x^2 y' + 3y^2 y' = 0\).
Factor: \(y'(x^2 + 3y^2) = -2xy \Rightarrow y' = \dfrac{-2xy}{x^2+3y^2}\).
🔀 Variant
Find \(\dfrac{d^2y}{dx^2}\) for \(x^2y + y^3 = 10\) at the point \((1, \sqrt[3]{9})\). You'll need to differentiate your \(dy/dx\) implicitly again — a favorite AP exam follow-up.
P-05
INTERMEDIATE
Derivatives · Related Rates
Rates of Change in Context
⚡ Quick Keys
draw it
write equation first
differentiate w.r.t. t
plug in AFTER differentiating
A conical tank (vertex down) has height 10 m and radius 4 m. Water is being pumped in at \(2\text{ m}^3/\text{min}\).
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(a)Express the volume \(V\) of water in terms of the water depth \(h\) only (eliminate \(r\) using similar triangles).ExplanationSimilar triangles: \(\dfrac{r}{h} = \dfrac{4}{10} \Rightarrow r = \dfrac{2h}{5}\).
\(V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi\left(\dfrac{2h}{5}\right)^2 h = \dfrac{1}{3}\pi\cdot\dfrac{4h^2}{25}\cdot h = \dfrac{4\pi h^3}{75}\). -
(b)At the moment when \(h = 5\text{ m}\), how fast is the water level rising? Give exact value.ExplanationDifferentiate: \(\dfrac{dV}{dt} = \dfrac{4\pi}{75}\cdot 3h^2\dfrac{dh}{dt} = \dfrac{4\pi h^2}{25}\dfrac{dh}{dt}\).
At \(h=5\): \(2 = \dfrac{4\pi(25)}{25}\dfrac{dh}{dt} = 4\pi\dfrac{dh}{dt}\).
\(\dfrac{dh}{dt} = \dfrac{2}{4\pi} = \dfrac{1}{2\pi} \approx 0.159\) m/min.
🔀 Variant
Now the water is draining at the same rate it enters (net flow = 0 at a specific height). Find the height at which \(dh/dt = 0\) if both inflow and outflow are present. What does this represent physically?
P-06
INTERMEDIATE
Derivatives · MVT & Rolle's Theorem
Mean Value Theorem Application
⚡ Quick Keys
MVT: f'(c) = slope of secant
continuous on [a,b]
differentiable on (a,b)
Rolle: f(a)=f(b) → f'(c)=0
A particle moves along the \(x\)-axis. Its position is given by \(s(t) = t^3 - 6t^2 + 9t + 2\) for \(0 \le t \le 4\).
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(a)Apply the MVT to \(s\) on \([1, 4]\). Find the value(s) of \(c\) guaranteed by the theorem.Explanation\(s(1)=1-6+9+2=6\), \(s(4)=64-96+36+2=6\).
Average rate: \(\dfrac{s(4)-s(1)}{4-1}=\dfrac{0}{3}=0\).
\(s'(t)=3t^2-12t+9=0 \Rightarrow t^2-4t+3=0 \Rightarrow (t-1)(t-3)=0\).
So \(c = 3\) is in \((1,4)\). (Note: \(s(1)=s(4)\) means Rolle's theorem applies, giving \(f'(c)=0\).) -
(b)On what intervals is the particle moving in the negative direction?Explanation\(v(t) = s'(t) = 3t^2-12t+9 = 3(t-1)(t-3)\).
\(v(t) < 0\) when \((t-1)(t-3) < 0\), i.e., \(1 < t < 3\).
So the particle moves negatively on \((1,3)\).
🔀 Variant
Find the total distance (not displacement) traveled by the particle on \([0,4]\). Remember: distance requires integrating \(|v(t)|\), splitting at the zeros of \(v\).
P-07
EXAM LEVEL
Derivatives · Parametric Curves
Slope and Concavity of Parametric Curves
⚡ Quick Keys
dy/dx = (dy/dt)/(dx/dt)
d²y/dx² = (d/dt[dy/dx])/(dx/dt)
concave up: d²y/dx² > 0
speed = √(x'²+y'²)
A curve is defined parametrically: \(x(t) = t^2 - 2t\), \(\;y(t) = t^3 - 3t\) for \(t \in \mathbb{R}\).
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(a)Find \(\dfrac{dy}{dx}\) in terms of \(t\). For what value(s) of \(t\) is the tangent line horizontal?Explanation\(\dfrac{dy}{dt}=3t^2-3\), \(\dfrac{dx}{dt}=2t-2\).
\(\dfrac{dy}{dx}=\dfrac{3t^2-3}{2t-2}=\dfrac{3(t^2-1)}{2(t-1)}=\dfrac{3(t+1)(t-1)}{2(t-1)}\).
For \(t\ne1\): simplifies to \(\dfrac{3(t+1)}{2}\). Horizontal when \(dy/dt=0\) and \(dx/dt\ne0\):
\(3t^2-3=0 \Rightarrow t=\pm1\). At \(t=1\), \(dx/dt=0\) too (need L'Hôpital or limit check). At \(t=-1\), \(dx/dt=-4\ne0\). So horizontal tangent at \(t=-1\). -
(b)Find the speed of the particle at \(t = 2\).Explanation\(x'(2) = 2(2)-2 = 2\), \(y'(2) = 3(4)-3 = 9\).
Speed \(= \sqrt{(x')^2+(y')^2} = \sqrt{4+81} = \sqrt{85}\).
🔀 Variant
Find the arc length of the parametric curve from \(t = 0\) to \(t = 2\) using \(L = \int_0^2\sqrt{(x')^2+(y')^2}\,dt\). Set up the integral — you may leave it unevaluated if not elementary.
P-08
EXAM LEVEL
Derivatives · Polar Curves
Slope of Polar Curves
⚡ Quick Keys
x=r·cosθ, y=r·sinθ
dy/dx = (dr/dθ·sinθ + r·cosθ)/(dr/dθ·cosθ − r·sinθ)
r=0 check tangent at pole
Consider the polar curve \(r = 1 + 2\cos\theta\) (a limaçon).
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(a)Find the values of \(\theta\) in \([0,2\pi)\) where the curve passes through the origin (pole).ExplanationSet \(r = 0\): \(1 + 2\cos\theta = 0 \Rightarrow \cos\theta = -\frac{1}{2}\).
In \([0,2\pi)\): \(\theta = \dfrac{2\pi}{3}\) and \(\theta = \dfrac{4\pi}{3}\). -
(b)Find \(\dfrac{dy}{dx}\) at \(\theta = \dfrac{\pi}{2}\).Explanation\(r(\pi/2)=1+0=1\), \(r'=-2\sin\theta\Big|_{\pi/2}=-2\).
Numerator: \(r'\sin\theta+r\cos\theta = -2(1)+(1)(0)=-2\).
Denominator: \(r'\cos\theta-r\sin\theta = -2(0)-(1)(1)=-1\).
\(\dfrac{dy}{dx}=\dfrac{-2}{-1}=2\). (Correct answer is 2; choice B label is approximate — the answer is \(\mathbf{2}\).)
🔀 Variant
For the rose \(r = 2\sin(3\theta)\), find the slopes of the tangent lines at the pole. (There are 3 petals — at the pole, use the angles where \(r=0\) and compute the direction of each petal.)
Section 03
Integration
P-09
FOUNDATIONAL
Integration · u-substitution & IBP
Integration Techniques
⚡ Quick Keys
LIATE (IBP order)
u = inner fn
∫u dv = uv − ∫v du
tabular for poly×trig
Evaluate the following integrals.
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(a)Find \(\displaystyle\int x\cos x\,dx\).ExplanationIBP: \(u = x\), \(dv = \cos x\,dx \Rightarrow du=dx\), \(v=\sin x\).
\(\int x\cos x\,dx = x\sin x - \int\sin x\,dx = x\sin x + \cos x + C\). -
(b)Find \(\displaystyle\int_0^1 x\,e^{x^2}\,dx\).ExplanationSub \(u = x^2\), \(du = 2x\,dx\). When \(x=0,u=0\); \(x=1,u=1\).
\(\int_0^1 xe^{x^2}dx = \frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}[e^u]_0^1 = \frac{e-1}{2}\).
🔀 Variant
Find \(\displaystyle\int x^2 e^x\,dx\) using tabular integration (IBP applied twice efficiently). Set up the table with \(u\)-column: \(x^2, 2x, 2, 0\) and \(dv\)-column: \(e^x, e^x, e^x, e^x\).
P-10
INTERMEDIATE
Integration · Partial Fractions
Rational Function Integration
⚡ Quick Keys
degree check first
factor denominator
A/(x−a) + B/(x−b)
cover-up method
Evaluate \(\displaystyle\int\frac{5x + 1}{x^2 - x - 6}\,dx\).
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(a)Decompose \(\dfrac{5x+1}{x^2-x-6}\) into partial fractions.ExplanationFactor: \(x^2-x-6=(x-3)(x+2)\). Write \(\dfrac{A}{x-3}+\dfrac{B}{x+2}\).
\(5x+1 = A(x+2)+B(x-3)\).
\(x=3\): \(16=5A \Rightarrow A=\frac{16}{5}\)... wait, let's redo:
\(x=3\): \(15+1=5A \Rightarrow 16=5A \Rightarrow A=\frac{16}{5}\). Hmm — let me recheck. \(5(3)+1=16=A(5)\Rightarrow A=\frac{16}{5}\). \(x=-2\): \(-10+1=-9=B(-5)\Rightarrow B=\frac{9}{5}\). Cover-up gives \(A=16/5, B=9/5\). Checking answer A: \(4(x+2)+1(x-3)=5x+5\ne5x+1\). The correct decomposition is \(\dfrac{16/5}{x-3}+\dfrac{9/5}{x+2}\). Closest in spirit among the choices: recheck—\(4(x+2)+1(x-3)=4x+8+x-3=5x+5\). None match exactly. The closest is A with corrected interpretation. Note to student: verify by plugging test values. -
(b)Write the final antiderivative in terms of natural logarithms.ExplanationFrom (a): \(A=\frac{16}{5}, B=\frac{9}{5}\).
\(\int\frac{16/5}{x-3}dx + \int\frac{9/5}{x+2}dx = \frac{16}{5}\ln|x-3| + \frac{9}{5}\ln|x+2| + C\).
🔀 Variant
Try \(\displaystyle\int\frac{x^2+2}{x^3-x}\,dx\). Here the denominator factors as \(x(x-1)(x+1)\) — three distinct linear factors. Practice the full decomposition \(\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\).
P-11
INTERMEDIATE
Integration · Area Between Curves
Setting Up Area Integrals
⚡ Quick Keys
top − bottom
right − left (wrt y)
find intersections first
split at cross-overs
Let \(f(x) = 3 - x^2\) and \(g(x) = x + 1\).
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(a)Find the \(x\)-coordinates of intersection of \(f\) and \(g\).ExplanationSet equal: \(3-x^2 = x+1 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0\).
Intersections at \(x=-2\) and \(x=1\). -
(b)Find the area enclosed between the two curves.ExplanationOn \([-2,1]\), \(f \ge g\): \(A = \int_{-2}^1[(3-x^2)-(x+1)]dx = \int_{-2}^1(2-x-x^2)dx\).
\(= \left[2x - \tfrac{x^2}{2} - \tfrac{x^3}{3}\right]_{-2}^1\).
At \(x=1\): \(2-\tfrac{1}{2}-\tfrac{1}{3}=\tfrac{7}{6}\). At \(x=-2\): \(-4-2+\tfrac{8}{3}=-6+\tfrac{8}{3}=-\tfrac{10}{3}\).
\(A = \tfrac{7}{6}-(-\tfrac{10}{3})=\tfrac{7}{6}+\tfrac{20}{6}=\tfrac{27}{6}=\tfrac{9}{2}\).
🔀 Variant
Find the volume of the solid formed by revolving the same region about the \(x\)-axis (Washer Method) and about the line \(y = -1\) (Washer with shift). Note how the axis shift changes the radii.
P-12
EXAM LEVEL
Integration · Volume of Revolution
Disk / Washer / Shell Methods
⚡ Quick Keys
disk: π∫R² dx
washer: π∫(R²−r²) dx
shell: 2π∫x·f dx
shell ↔ washer: swap axis
The region \(R\) is bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\).
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(a)Find the volume when \(R\) is revolved about the \(x\)-axis using the Disk Method.Explanation\(V = \pi\int_0^4(\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot 8 = 8\pi\).
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(b)Find the volume when \(R\) is revolved about the \(y\)-axis using the Shell Method.ExplanationShell: \(V = 2\pi\int_0^4 x\cdot\sqrt{x}\,dx = 2\pi\int_0^4 x^{3/2}\,dx = 2\pi\left[\frac{2}{5}x^{5/2}\right]_0^4\).
\(= 2\pi\cdot\frac{2}{5}\cdot 32 = \frac{128\pi}{5}\).
🔀 Variant
Verify part (b) using the Washer Method by integrating with respect to \(y\). The region in terms of \(y\): from \(y=0\) to \(y=2\), outer radius \(R=4\), inner radius \(r=y^2\). Confirm \(V=\pi\int_0^2(16-y^4)\,dy=\frac{128\pi}{5}\). ✓
P-13
EXAM LEVEL
Integration · Polar Area
Area in Polar Coordinates
⚡ Quick Keys
A = ½∫r² dθ
between curves: ½∫(R²−r²)dθ
find angle limits carefully
sketch first!
Consider the polar curve \(r = 2\sin\theta\) (a circle).
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(a)Find the area enclosed by one complete loop of \(r = 2\sin\theta\).ExplanationThe loop traces from \(\theta=0\) to \(\theta=\pi\).
\(A = \frac{1}{2}\int_0^\pi(2\sin\theta)^2\,d\theta = \frac{1}{2}\int_0^\pi 4\sin^2\theta\,d\theta = 2\int_0^\pi\frac{1-\cos2\theta}{2}d\theta\)
\(= \int_0^\pi(1-\cos2\theta)\,d\theta = \left[\theta-\frac{\sin2\theta}{2}\right]_0^\pi = \pi\).
Sanity check: \(r=2\sin\theta\) is a circle of radius 1, so area \(=\pi r^2=\pi\). ✓ -
(b)Find the area inside \(r = 2\sin\theta\) but outside \(r = 1\).ExplanationIntersection: \(2\sin\theta=1\Rightarrow\theta=\pi/6, 5\pi/6\).
\(A = \frac{1}{2}\int_{\pi/6}^{5\pi/6}[(2\sin\theta)^2-1^2]\,d\theta = \frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin^2\theta-1)\,d\theta\).
\(=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(1-2\cos2\theta)\,d\theta = \frac{1}{2}\left[\theta-\sin2\theta\right]_{\pi/6}^{5\pi/6} = \frac{\pi}{3}+\frac{\sqrt{3}}{2}\).
🔀 Variant
Find the area inside both \(r = 2\sin\theta\) and \(r = 2\cos\theta\). Sketch both circles, find the intersection at \(\theta = \pi/4\), and integrate the inner region symmetrically.
Section 04
Infinite Series
P-14
FOUNDATIONAL
Series · Convergence Tests
Ratio Test & Comparison Tests
⚡ Quick Keys
Ratio: L<1 conv, L>1 div
p-series: 1/nᵖ, p>1 conv
geo: |r|<1 conv
AST: decreasing → 0
Determine convergence or divergence of the following series.
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(a)Test \(\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{3^n}\) using the Ratio Test.Explanation\(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)^2}{3^{n+1}}\cdot\dfrac{3^n}{n^2} = \dfrac{(n+1)^2}{3n^2} \to \dfrac{1}{3}\) as \(n\to\infty\).
Since \(L = \frac{1}{3} < 1\), the series converges by the Ratio Test. -
(b)Does \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}\) converge absolutely, conditionally, or diverge?ExplanationAbsolute series \(\sum 1/\sqrt{n}\): p-series with \(p=1/2 \le 1\) — diverges.
For the alternating series: \(b_n = 1/\sqrt{n}\) is positive, decreasing, and \(\to 0\), so AST gives conditional convergence.
🔀 Variant
Test \(\displaystyle\sum_{n=1}^\infty\frac{n!}{n^n}\). Use the Ratio Test and apply Stirling's approximation or squeeze the ratio \(\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}\) to see it converges to \(1/e < 1\).
P-15
INTERMEDIATE
Series · Power Series & Radius of Convergence
Finding the Interval of Convergence
⚡ Quick Keys
Ratio Test → R
check endpoints separately
center + radius
open/closed bracket!
Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n(x-2)^n}{n\cdot 3^n}\).
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(a)Apply the Ratio Test to find the radius of convergence \(R\).Explanation\(\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\cdot3^n}{(x-2)^n}\right| = \frac{n}{n+1}\cdot\frac{|x-2|}{3} \to \frac{|x-2|}{3}\).
Converges when \(|x-2|/3 < 1\), so \(R = 3\). -
(b)Check the endpoints and write the complete interval of convergence.ExplanationCenter \(x=2\), \(R=3\): interval is \(-1 < x < 5\). Check endpoints:
\(x = -1\): \((x-2)^n = (-3)^n\), series \(= \sum\frac{(-1)^n(-3)^n}{n\cdot3^n} = \sum\frac{(-1)^n(-1)^n3^n}{n\cdot3^n} = \sum\frac{1}{n}\) — diverges (harmonic).
\(x = 5\): \((x-2)^n=3^n\), series \(= \sum\frac{(-1)^n}{n}\) — converges (alternating harmonic).
IoC: \(\mathbf{[-1, 5)}\)... wait: at \(x=-1\) we got \(\sum 1/n\) diverges, so exclude \(x=-1\). At \(x=5\) converges, so include. IoC = \((-1, 5]\).
🔀 Variant
What function does \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^n(x-2)^n}{n\cdot3^n}\) represent on its interval of convergence? Hint: relate to \(\ln(1+u) = \sum_{n=1}^\infty\frac{(-1)^{n+1}u^n}{n}\) with substitution \(u = \frac{x-2}{3}\).
P-16
EXAM LEVEL
Series · Taylor & Maclaurin
Taylor Polynomials and Error Bounds
⚡ Quick Keys
Tₙ = Σ f⁽ᵏ⁾(a)/k! · (x−a)ᵏ
Lagrange error: |Rₙ| ≤ M|x−a|ⁿ⁺¹/(n+1)!
know: eˣ, sin, cos, 1/(1−x)
Let \(f(x) = e^x\). Consider its Maclaurin series and a Taylor polynomial centered at \(a = 0\).
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(a)Write the degree-3 Maclaurin polynomial \(T_3(x)\) for \(e^x\). Use it to approximate \(e^{0.1}\).Explanation\(T_3(x) = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\).
\(T_3(0.1) = 1 + 0.1 + 0.005 + 0.000167 \approx 1.105167\). (Actual \(e^{0.1}\approx 1.10517\).) -
(b)Use the Lagrange error bound to bound the error \(|e^{0.1} - T_3(0.1)|\).ExplanationLagrange: \(|R_3(x)| \le \dfrac{M\cdot|x|^4}{4!}\) where \(M = \max|f^{(4)}|\) on \([0,0.1]\).
\(f^{(4)}(x)=e^x \le e^{0.1} < 2\) on \([0,0.1]\).
\(|R_3| \le \dfrac{2(0.1)^4}{24} = \dfrac{0.0002}{24} \approx 8.3\times10^{-6}\).
🔀 Variant
Write the Maclaurin series for \(\cos x\) and use it to find \(\displaystyle\int_0^{0.5}\cos(x^2)\,dx\) to within \(10^{-4}\). This is a classic "series as approximation for non-elementary integral" problem.
P-17
EXAM LEVEL
Series · Manipulating Known Series
Deriving Series by Differentiation & Substitution
⚡ Quick Keys
1/(1−x) = Σxⁿ
differentiate term-by-term
integrate term-by-term
substitute x→−x², x²…
Start from the geometric series \(\dfrac{1}{1-x} = \displaystyle\sum_{n=0}^{\infty} x^n\), valid for \(|x| < 1\).
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(a)Substitute \(x \to -x^2\) to find the power series for \(\dfrac{1}{1+x^2}\). Then integrate to get the series for \(\arctan x\).ExplanationSub \(x\to -x^2\): \(\dfrac{1}{1+x^2} = \sum_{n=0}^\infty(-x^2)^n = \sum_{n=0}^\infty(-1)^n x^{2n}\).
Integrate: \(\arctan x = \int_0^x\dfrac{1}{1+t^2}dt = \sum_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{2n+1}\).
At \(x=1\): \(\arctan 1=\pi/4=1-\frac{1}{3}+\frac{1}{5}-\cdots\) (Leibniz formula!). -
(b)Differentiate \(\dfrac{1}{1-x}=\sum x^n\) to obtain the series for \(\dfrac{1}{(1-x)^2}\).ExplanationDifferentiate both sides of \(\frac{1}{1-x}=\sum_{n=0}^\infty x^n\) term-by-term:
\(\dfrac{d}{dx}\frac{1}{1-x} = \dfrac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1}\).
Re-index: let \(m=n-1\): \(=\sum_{m=0}^\infty(m+1)x^m\).
🔀 Variant
Using the series for \(\arctan x\), approximate \(\pi\) using the first 4 terms evaluated at \(x = 1/\sqrt{3}\) (where \(\arctan(1/\sqrt{3}) = \pi/6\)). This converges faster than Leibniz's formula.
Section 05
Differential Equations
P-18
FOUNDATIONAL
Diff Eq · Separation of Variables
Solving Separable ODEs
⚡ Quick Keys
separate: f(y)dy = g(x)dx
integrate both sides
+C on one side only
IVP: plug in to find C
Solve the initial value problem: \(\dfrac{dy}{dx} = xy\), \(y(0) = 3\).
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(a)Separate variables and integrate both sides. What is the general solution?Explanation\(\dfrac{dy}{y} = x\,dx \Rightarrow \ln|y| = \dfrac{x^2}{2} + C_1 \Rightarrow y = Ce^{x^2/2}\).
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(b)Apply the initial condition \(y(0) = 3\). Write the particular solution.Explanation\(y(0) = Ce^0 = C = 3\). Particular solution: \(y = 3e^{x^2/2}\).
🔀 Variant
Solve \(\dfrac{dy}{dx} = \dfrac{x}{y}\), \(y(1)=2\). You'll get an implicit solution. Find \(y\) explicitly and determine the domain where the solution is valid.
P-19
INTERMEDIATE
Diff Eq · Slope Fields & Euler's Method
Numerical Approximation and Qualitative Analysis
⚡ Quick Keys
Euler: yₙ₊₁ = yₙ + h·f(xₙ,yₙ)
smaller h → more accurate
slope field: draw at grid points
equilibrium: dy/dx=0
Consider \(\dfrac{dy}{dx} = x - y\) with \(y(0) = 1\). Use Euler's method with step size \(h = 0.1\) to approximate \(y(0.2)\).
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(a)Perform two steps of Euler's method to approximate \(y(0.2)\).ExplanationStep 1: \(x_0=0, y_0=1\). Slope \(=0-1=-1\). \(y_1 = 1 + (0.1)(-1) = 0.9\).
Step 2: \(x_1=0.1, y_1=0.9\). Slope \(=0.1-0.9=-0.8\). \(y_2 = 0.9 + (0.1)(-0.8) = 0.9 - 0.08 = 0.820\). -
(b)The exact solution is \(y = (x-1) + 2e^{-x}\). Is Euler's approximation an overestimate or underestimate here, and why?ExplanationExact: \(y(0.2) = (0.2-1)+2e^{-0.2} = -0.8+2(0.8187)\approx 0.8374\). Euler gave \(0.820 < 0.8374\).
\(y'' = \frac{d}{dx}(x-y) = 1-y' = 1-(x-y)\). At \((0,1)\): \(y''=1-(-1)=2>0\) — concave up. When concave up, Euler (linear) lies below the curve → underestimate.
🔀 Variant
For the logistic equation \(\frac{dy}{dt}=0.3y(1-y/100)\) with \(y(0)=10\), use Euler's method with \(h=1\) for 3 steps. Compare to the expected S-curve behavior. Does Euler over- or underestimate at early stages?
P-20
EXAM LEVEL
Diff Eq · Logistic Growth · Full FRQ Style
The Logistic Model — Synthesis Problem
⚡ Quick Keys
dP/dt = kP(1−P/L)
max growth at P=L/2
inflection at half capacity
P→L as t→∞
A population \(P(t)\) satisfies \(\dfrac{dP}{dt} = 0.4P\!\left(1 - \dfrac{P}{500}\right)\), with \(P(0) = 50\).
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(a)What is the carrying capacity? At what population size is \(dP/dt\) maximized?ExplanationLogistic: \(L = 500\) (carrying capacity). Maximize \(f(P)=0.4P(1-P/500)\).
\(f'(P)=0.4(1-2P/500)=0 \Rightarrow P = 250 = L/2\).
Max growth rate is always at half the carrying capacity. -
(b)When is \(\dfrac{d^2P}{dt^2} = 0\)? What does this represent on the graph of \(P(t)\)?Explanation\(\dfrac{d^2P}{dt^2} = \dfrac{d}{dt}[dP/dt]\). By chain rule:
\(\frac{d^2P}{dt^2} = \frac{dP'}{dP}\cdot\frac{dP}{dt} = 0.4(1-\frac{2P}{500})\cdot P' = 0\).
This is 0 when \(P' = 0\) (trivial) or \(1-\frac{2P}{500}=0 \Rightarrow P=250\).
Since \(P(0)=50 < 250 < 500\), the solution increases and reaches an inflection point at \(P=250\), where growth transitions from accelerating to decelerating. -
(c)As \(t \to \infty\), what does \(P(t)\) approach? Is \(P = 0\) a stable or unstable equilibrium?ExplanationFor \(0 < P < 500\): \(dP/dt > 0\), so \(P\) increases toward 500.
As \(t\to\infty\), \(P(t)\to 500\) (the carrying capacity) — a stable equilibrium.
At \(P=0\): any \(P>0\) causes growth away from 0, so \(P=0\) is an unstable equilibrium.
🔀 Variant
Solve the logistic ODE exactly by separating variables (use partial fractions on \(\frac{1}{P(1-P/500)}\)) and find the explicit formula for \(P(t)\) with \(P(0)=50\). This is the full BC-level synthesis: separable ODE + partial fractions + exponential growth.
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