Unit 1 · Limits & Continuity
01
L'HÔPITAL = 0/0 or ∞/∞ ONLY
Evaluate the limit:
\[\lim_{x \to 0} \frac{\sin(3x)}{5x}\]
\[\lim_{x \to 0} \frac{\sin(3x)}{5x}\]
⚠ Trap: Students often write 3/5 without justification — know why.
Explanation
Using the fundamental limit \(\lim_{x\to0}\frac{\sin u}{u}=1\), rewrite: \(\frac{\sin(3x)}{5x} = \frac{3}{5}\cdot\frac{\sin(3x)}{3x}\). As \(x\to 0\), \(\frac{\sin(3x)}{3x}\to 1\), so the limit is \(\dfrac{3}{5}\). L'Hôpital also works (0/0 form): \(\frac{3\cos(3x)}{5}\big|_{x=0}=\frac{3}{5}\).
02
CONTINUITY: Exist · Equal · Defined
Let \(f(x)=\begin{cases} \dfrac{x^2-4}{x-2} & x\neq 2 \\ k & x=2 \end{cases}\). For what value of \(k\) is \(f\) continuous at \(x=2\)?
⚠ Trap: The numerator factors — don't just plug in 2.
Explanation
Factor: \(\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2\) for \(x\neq 2\). So \(\lim_{x\to2}f(x)=4\). For continuity, \(f(2)=k\) must equal this limit: \(k=4\).
Unit 2 · Differentiation
03
CHAIN RULE: outside → inside, multiply by derivative of inside
If \(f(x) = e^{\sin(x^2)}\), find \(f'(x)\).
⚠ Trap: Three layers of composition — apply chain rule twice.
Explanation
Three layers: outer \(e^u\), middle \(\sin v\), inner \(x^2\).\(f'(x)= e^{\sin(x^2)}\cdot \cos(x^2)\cdot 2x\). Each derivative multiplies from outside in.
04
IMPLICIT DIFF: treat y as function, chain rule on y-terms, solve for dy/dx
For the curve \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\) at the point \((3, 4)\).
Explanation
Differentiate both sides: \(2x + 2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x}{y}\). At \((3,4)\): \(-\frac{3}{4}\). This makes sense geometrically — the circle's tangent at \((3,4)\) slopes downward.
05
PRODUCT RULE: (uv)' = u'v + uv'
Let \(g(x) = x^3 \ln x\). Find \(g''(x)\).
⚠ Trap: You must differentiate twice. Don't stop at \(g'(x)\).
Explanation
\(g'(x)=3x^2\ln x + x^3\cdot\frac{1}{x}=3x^2\ln x+x^2\).\(g''(x)=6x\ln x+3x^2\cdot\frac{1}{x}+2x=6x\ln x+3x+2x=6x\ln x+5x\).
Unit 3 · Applications of Derivatives
06
MVT: f'(c) = [f(b)-f(a)]/(b-a), need continuous & differentiable
The function \(f(x)=x^3-x\) on \([0,2]\). The Mean Value Theorem guarantees a \(c\in(0,2)\) with \(f'(c)=?\)
Explanation
MVT gives the average rate of change: \(\frac{f(2)-f(0)}{2-0}=\frac{(8-2)-0}{2}=\frac{6}{2}=3\). So \(f'(c)=3\). (Verify: \(f'(x)=3x^2-1=3\Rightarrow x=\pm\frac{2}{\sqrt{3}}\). The positive root lies in \((0,2)\). ✓)
07
CONCAVITY: f'' > 0 → concave up (smile), f'' < 0 → concave down (frown)
Let \(f(x)=x^4 - 4x^3\). Where does the graph of \(f\) have an inflection point?
⚠ Trap: An inflection point requires \(f''\) to change sign, not just equal zero.
Explanation
\(f''(x)=12x^2-24x=12x(x-2)\). Zeros at \(x=0\) and \(x=2\). Sign test: on \((-\infty,0)\), \(f''>0\); on \((0,2)\), \(f''<0\); on \((2,\infty)\), \(f''>0\). Sign changes at both! But check \(x=0\): wait — sign goes from \(+\) to \(-\) → inflection. And \(x=2\): sign goes from \(-\) to \(+\) → inflection. Both are inflection points. (Answer C is actually also correct — be careful with the options. The "only \(x=2\)" choice is the trap!)
Note: Both \(x=0\) and \(x=2\) are inflection points. The correct answer is C.
08
RELATED RATES: draw diagram → label variables → differentiate with respect to t
A ladder 10 ft long leans against a wall. The bottom slides away at 2 ft/sec. How fast is the top sliding down when the bottom is 6 ft from the wall?
Explanation
Let \(x\) = base, \(y\) = height. Constraint: \(x^2+y^2=100\). Differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). When \(x=6\): \(y=\sqrt{100-36}=8\). Plug in \(\frac{dx}{dt}=2\): \(2(6)(2)+2(8)\frac{dy}{dt}=0\Rightarrow\frac{dy}{dt}=-\frac{3}{2}\) ft/sec. Speed = 1.5 ft/sec downward.
Unit 4 · Integration
09
u-SUB: pick u = inside function, rewrite everything in u, back-substitute
Evaluate \(\displaystyle\int x\sqrt{x^2+1}\,dx\).
Explanation
Let \(u=x^2+1\), \(du=2x\,dx\), so \(x\,dx=\frac{du}{2}\). Integral becomes \(\int\sqrt{u}\cdot\frac{du}{2}=\frac{1}{2}\cdot\frac{2}{3}u^{3/2}+C=\frac{1}{3}(x^2+1)^{3/2}+C\).
10
FTC Part 1: d/dx ∫[a to g(x)] f(t)dt = f(g(x))·g'(x)
Let \(F(x)=\displaystyle\int_1^{x^2} \sqrt{t+1}\,dt\). Find \(F'(x)\).
⚠ Trap: Upper limit is \(x^2\), not \(x\) — chain rule is required!
Explanation
FTC Part 1 with chain rule: \(F'(x)=\sqrt{(x^2)+1}\cdot\frac{d}{dx}(x^2)=\sqrt{x^2+1}\cdot 2x\). The key mistake is forgetting to multiply by \(\frac{d}{dx}(x^2)=2x\).
11
IBP: ∫udv = uv − ∫vdu · LIATE order: Logs, Inverse trig, Algebraic, Trig, Exponential
Evaluate \(\displaystyle\int x\,e^x\,dx\).
Explanation
IBP: let \(u=x\), \(dv=e^x\,dx\), so \(du=dx\), \(v=e^x\).\(\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x+C=e^x(x-1)+C\).
Unit 5 · Infinite Series
12
RATIO TEST: L = lim|a(n+1)/a(n)| · L<1 converges · L>1 diverges · L=1 inconclusive
Does the series \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}\) converge or diverge?
Explanation
Ratio test: \(\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^n=\frac{1}{(1+\frac{1}{n})^n}\to\frac{1}{e}<1\). Converges.
13
TAYLOR: f(x) = Σ f⁽ⁿ⁾(a)/n! · (x-a)ⁿ · Memorize e^x, sin x, cos x, 1/(1-x)
The Maclaurin series for \(\sin x\) is \(x - \dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\). What is the Maclaurin series for \(\sin(x^2)\)?
Explanation
Simply substitute \(x^2\) for \(x\) in the series: \(\sin(x^2)=x^2-\frac{(x^2)^3}{3!}+\frac{(x^2)^5}{5!}-\cdots=x^2-\frac{x^6}{6}+\frac{x^{10}}{120}-\cdots\).
14
P-SERIES: Σ1/nᵖ converges iff p > 1
Which of the following series diverges?
⚠ Trap: \(p=1\) is the harmonic series — it diverges even though terms go to 0.
Explanation
\(\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}\) is a p-series with \(p=\frac{1}{2}<1\), so it diverges. All others have \(p>1\): A has \(p=2\), B has \(p=1.5\), D has \(p=1.01\) — all converge.
Unit 6 · Parametric & Polar
15
PARAMETRIC SLOPE: dy/dx = (dy/dt)/(dx/dt)
A curve is defined parametrically by \(x(t)=t^2\) and \(y(t)=t^3-3t\). Find all values of \(t\) where the tangent line is horizontal.
Explanation
Horizontal tangent when \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\neq 0\). \(\frac{dy}{dt}=3t^2-3=3(t^2-1)=0\Rightarrow t=\pm1\). Check \(\frac{dx}{dt}=2t\): at \(t=\pm1\), \(2t=\pm2\neq0\). Both valid.
16
ARC LENGTH (param): L = ∫√[(dx/dt)² + (dy/dt)²] dt
The arc length of the curve \(x=3t^2,\; y=4t^2\) from \(t=0\) to \(t=1\) equals:
⚠ Trap: This simplifies nicely — factor before integrating.
Explanation
\(\frac{dx}{dt}=6t\), \(\frac{dy}{dt}=8t\). \(\sqrt{(6t)^2+(8t)^2}=\sqrt{36t^2+64t^2}=\sqrt{100t^2}=10t\) (for \(t\geq0\)). \(L=\int_0^1 10t\,dt=5t^2\big|_0^1=5\). Wait — answer is 5. The correct answer is A.
Note: \(L = 5\). (Answer A is correct.)
17
POLAR AREA: A = ½∫r² dθ between θ₁ and θ₂
Find the area enclosed by one petal of the rose \(r=\cos(3\theta)\).
Explanation
One petal of \(r=\cos(3\theta)\) spans \(\theta\in[-\pi/6,\,\pi/6]\). Area \(=\frac{1}{2}\int_{-\pi/6}^{\pi/6}\cos^2(3\theta)\,d\theta=\frac{1}{4}\int_{-\pi/6}^{\pi/6}(1+\cos6\theta)\,d\theta=\frac{1}{4}\left[\theta+\frac{\sin6\theta}{6}\right]_{-\pi/6}^{\pi/6}=\frac{1}{4}\cdot\frac{\pi}{3}=\frac{\pi}{12}\).
Unit 7 · Differential Equations
18
SEP. OF VARIABLES: group y-terms one side, x-terms other → integrate both sides
Solve the differential equation \(\dfrac{dy}{dx} = \dfrac{x}{y}\) with \(y(0)=3\).
Explanation
Separate: \(y\,dy = x\,dx\). Integrate: \(\frac{y^2}{2}=\frac{x^2}{2}+C\). Apply \(y(0)=3\): \(\frac{9}{2}=C\). So \(y^2=x^2+9\), thus \(y=\sqrt{x^2+9}\) (positive since \(y(0)=3>0\)).
19
LOGISTIC: dP/dt = kP(M−P) · carrying capacity = M · max growth at P = M/2
A population follows the logistic model \(\dfrac{dP}{dt}=0.4P\!\left(1-\dfrac{P}{500}\right)\). The population grows most rapidly when:
Explanation
The logistic rate \(f(P)=kP(M-P)\) is a downward parabola in \(P\), maximized at \(P=\frac{M}{2}=\frac{500}{2}=250\). At \(P=500\), growth stops (carrying capacity). At \(P=0\), the rate is also 0.
20
EULER'S METHOD: y(new) = y(old) + h·f(x,y) · small h = more accurate
Use Euler's method with step size \(h=0.5\) to approximate \(y(1)\) given \(\dfrac{dy}{dx}=x+y\) and \(y(0)=1\).
⚠ Trap: Two steps needed — from \(x=0\) to \(x=0.5\) to \(x=1\).
Explanation
Step 1: At \((0,1)\), slope \(=0+1=1\). New \(y=1+0.5(1)=1.5\).Step 2: At \((0.5,\,1.5)\), slope \(=0.5+1.5=2\). New \(y=1.5+0.5(2)=2.5\).
Wait — \(y(1)\approx2.5\). Check D. (Answer is D: 2.50.)
Note: The correct numerical answer is \(2.5\), so choose D.