TRAP: 0/0 ≠ no limit — always FACTOR first
Evaluate the limit:
\[\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\]
Quick Example
Factor \(x^2 - 9 = (x-3)(x+3)\), then cancel \((x-3)\). You get \(\lim_{x\to 3}(x+3) = 6\).
A \(0\)B \(6\)C DNE (does not exist)D \(9\)
CONTINUITY CHECK: f(a) exists · limit exists · they're EQUAL
The function \(f(x) = \dfrac{x^2 - 4}{x - 2}\) is not continuous at \(x = 2\) because:
A The limit does not exist at \(x = 2\)B \(f(2)\) is negativeC \(f(2)\) is undefined (hole at \(x=2\))D The derivative is undefined
SQUEEZE: sin(θ)/θ → 1 as θ → 0 (radians only!)
Evaluate:
\[\lim_{x \to 0} \frac{\sin(5x)}{x}\]
Trick
Multiply top and bottom by 5: \(\dfrac{\sin(5x)}{5x} \cdot 5\). As \(x\to 0\), \(\dfrac{\sin(5x)}{5x}\to 1\), so the answer is \(5\).
A \(0\)B \(1\)C \(5\)D DNE
IVT: f continuous on [a,b] → hits EVERY value between f(a) and f(b)
Let \(f\) be continuous on \([1, 4]\) with \(f(1) = -2\) and \(f(4) = 7\). By the Intermediate Value Theorem , which of the following is guaranteed?
A There exists \(c \in (1,4)\) such that \(f(c) = 0\)B \(f\) is increasing on \([1,4]\)C \(f\) has a maximum at some \(c \in [1,4]\)D The average rate of change is \(\frac{9}{3} = 3\)
CHAIN RULE: d/dx[f(g(x))] = f'(g(x)) · g'(x) — "outer × inner'"
Find \(\dfrac{d}{dx}\left[\sin(x^3)\right]\).
Step-by-step
Outer function: \(\sin(\square)\), so outer derivative = \(\cos(\square)\).
A \(\cos(x^3)\)B \(3x^2\cos(x^3)\)C \(3x^2\sin(x^3)\)D \(-3x^2\cos(x^3)\)
PRODUCT RULE: (uv)' = u'v + uv' — don't just multiply derivatives!
Find \(\dfrac{d}{dx}\left[x^2 \cdot e^x\right]\).
A \(2x \cdot e^x\)B \(x^2 e^x\)C \(2xe^x + x^2 e^x\)D \(2x + e^x\)
IMPLICIT DIFF: treat y as y(x), apply chain rule → dy/dx appears, then SOLVE for it
Given \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\) using implicit differentiation.
Process
Differentiate both sides: \(2x + 2y\dfrac{dy}{dx} = 0\). Solve: \(\dfrac{dy}{dx} = -\dfrac{x}{y}\).
A \(\dfrac{x}{y}\)B \(2x + 2y\)C \(-\dfrac{y}{x}\)D \(-\dfrac{x}{y}\)
TANGENT LINE: y − f(a) = f'(a)(x − a) — plug in the POINT then the SLOPE
The line tangent to \(f(x) = x^3 - 2x\) at \(x = 1\) has equation:
A \(y = x - 2\)B \(y = 3x - 4\)C \(y = -x\)D \(y = x + 2\)
MVT: f'(c) = [f(b)−f(a)]/(b−a) — average slope = instantaneous slope somewhere
Let \(f(x) = x^2\) on \([1, 3]\). By the Mean Value Theorem , the value of \(c \in (1,3)\) satisfying \(f'(c) = \dfrac{f(3)-f(1)}{3-1}\) is:
A \(c = 1\)B \(c = 2\)C \(c = \sqrt{3}\)D \(c = 3\)
RELATED RATES: draw diagram · write equation · DIFFERENTIATE with respect to t
A ladder 10 ft long leans against a wall. The bottom slides away at 2 ft/s. When the bottom is 6 ft from the wall, how fast is the top sliding down ?
Setup
\(x^2 + y^2 = 100\). Differentiate: \(2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}=0\).
A \(\dfrac{3}{4}\) ft/sB \(2\) ft/sC \(\dfrac{3}{2}\) ft/sD \(\dfrac{4}{3}\) ft/s
1ST DERIV TEST: f' + → − means LOCAL MAX; f' − → + means LOCAL MIN
Let \(f'(x) = (x-1)(x-4)\). Which of the following is true about local extrema of \(f\)?
A \(f\) has a local max at \(x=1\) and local min at \(x=4\)B \(f\) has local minima at both \(x=1\) and \(x=4\)C \(f\) has no local extremaD \(f\) has a local min at \(x=1\) and local max at \(x=4\)
CONCAVITY: f'' > 0 → concave UP (smile); f'' < 0 → concave DOWN (frown)
The function \(f(x) = x^4 - 4x^3\). On what interval is \(f\) concave up ?
Process
\(f'(x) = 4x^3-12x^2\). \(f''(x) = 12x^2-24x = 12x(x-2)\).
A \((0, 2)\) onlyB \((-\infty, 0) \cup (2, \infty)\)C \((2, \infty)\) onlyD \((-\infty, 2)\)
OPTIMIZE: set f'(x)=0 → find critical pts → verify with 1st or 2nd deriv test
A rectangle has perimeter 40. What dimensions maximize its area?
A 10 × 10 (square)B 5 × 15C 2 × 18D 8 × 12
POWER RULE (∫): ∫x^n dx = x^(n+1)/(n+1) + C — add 1 to exponent, divide by new exponent
Evaluate \(\displaystyle\int (3x^2 + 2x - 5)\, dx\).
A \(x^3 + x^2 - 5x + C\)B \(6x + 2 + C\)C \(3x^3 + 2x^2 - 5x + C\)D \(x^3 + x^2 + C\)
FTC Part 1: d/dx[∫(a to x) f(t)dt] = f(x) — just PLUG IN x, don't integrate!
If \(F(x) = \displaystyle\int_1^x (t^2 + 1)\,dt\), then \(F'(x) =\)
A \(\dfrac{x^3}{3} + x - \dfrac{4}{3}\)B \(x^2 + 1\)C \(2x\)D \(t^2 + 1\)
FTC Part 2: ∫(a to b)f(x)dx = F(b) − F(a) — evaluate antiderivative at BOUNDS
Evaluate \(\displaystyle\int_0^2 (x^2 - x)\,dx\).
A \(4\)B \(0\)C \(\dfrac{2}{3}\)D \(2\)
U-SUB: pick u = inner function → find du → REPLACE everything, no x's left!
Evaluate \(\displaystyle\int 2x\cos(x^2)\,dx\).
U-substitution
Let \(u = x^2\), then \(du = 2x\,dx\). Integral becomes \(\int \cos(u)\,du = \sin(u) + C = \sin(x^2)+C\).
A \(-2x\sin(x^2) + C\)B \(\cos(x^2) + C\)C \(2\sin(x^2) + C\)D \(\sin(x^2) + C\)
AREA BETWEEN CURVES: ∫[TOP − BOTTOM]dx — always subtract the LOWER function
Find the area between \(f(x) = x^2\) and \(g(x) = x\) on \([0, 1]\).
Which is on top?
On \([0,1]\): \(x \geq x^2\), so \(\displaystyle\int_0^1(x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \frac{1}{6}\).
A \(\dfrac{1}{6}\)B \(\dfrac{1}{3}\)C \(\dfrac{1}{2}\)D \(1\)
SEPARABLE DE: move y's to one side, x's to other → INTEGRATE both sides → solve for y
Solve the differential equation \(\dfrac{dy}{dx} = 2xy\) with \(y(0) = 3\).
Steps
Separate: \(\dfrac{dy}{y} = 2x\,dx\). Integrate: \(\ln|y| = x^2 + C\). Exponentiate: \(y = Ae^{x^2}\). Apply IC: \(A = 3\).
A \(y = e^{x^2}\)B \(y = 3e^{2x}\)C \(y = 3e^{x^2}\)D \(y = x^2 + 3\)
POSITION / VELOCITY / ACCELERATION: v=s', a=v'=s'' · displacement = ∫v dt
A particle moves along the \(x\)-axis with velocity \(v(t) = t^2 - 4\). What is the total distance (not displacement) traveled on \([0, 3]\)?
Key Distinction
Total distance = \(\displaystyle\int_0^3|v(t)|\,dt\). The particle reverses at \(t=2\) (where \(v=0\)).
A \(3\)B \(7\)C \(\dfrac{16}{3}\)D \(-3\)
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