Exhaustive Calculus Practice Problems

Category 1: Limits & Asymptotes

Problem 1.1: The Factoring Trap (Beginner)
Evaluate: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

• Proof Step 1: Direct substitution yields $\frac{9-9}{3-3} = \frac{0}{0}$. Indeterminate form.
• Proof Step 2: Factor the numerator using difference of squares: $(x-3)(x+3)$.
• Proof Step 3: The expression is now $\frac{(x-3)(x+3)}{x-3}$. Cancel the $(x-3)$ terms.
• Proof Step 4: You are left with $\lim_{x \to 3} (x + 3)$.
• Final Answer: Substitute $3$ into the simplified expression: $3 + 3 = 6$.

Problem 1.2: Infinity and Rational Functions (Intermediate)
Evaluate: $\lim_{x \to \infty} \frac{4x^3 - 2x^2 + 7}{5x^3 + x - 1}$

• Proof Step 1: Both numerator and denominator go to $\infty$. We look at the leading coefficients.
• Proof Step 2: The highest power of $x$ in the entire fraction is $x^3$. Divide every single term by $x^3$.
• Proof Step 3: $\frac{4 - \frac{2}{x} + \frac{7}{x^3}}{5 + \frac{1}{x^2} - \frac{1}{x^3}}$.
• Proof Step 4: As $x$ approaches infinity, any constant divided by $x$ approaches zero.
• Final Answer: The limit evaluates exactly to the ratio of the leading coefficients: $\frac{4}{5}$.

Problem 1.3: Advanced L'Hôpital's Application (Advanced)
Evaluate: $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

• Proof Step 1: Direct substitution gives $\frac{1 - 1 - 0}{0} = \frac{0}{0}$. Apply L'Hôpital's rule.
• Proof Step 2: Derivative of numerator is $e^x - 1$. Derivative of denominator is $2x$. The new limit is $\lim_{x \to 0} \frac{e^x - 1}{2x}$.
• Proof Step 3: Substitute 0 again. It yields $\frac{1-1}{0} = \frac{0}{0}$. We must apply L'Hôpital's rule a second time.
• Proof Step 4: Second derivative of numerator is $e^x$. Second derivative of denominator is $2$. The limit is now $\lim_{x \to 0} \frac{e^x}{2}$.
• Final Answer: Plug in $0$. $e^0 = 1$. The final limit is $\frac{1}{2}$.

Problem 1.4: The Squeeze Theorem (Intermediate)
Evaluate: $\lim_{x \to 0} x^2 \sin(\frac{1}{x})$

• Proof Step 1: We know that $-1 \leq \sin(\frac{1}{x}) \leq 1$ for all $x \neq 0$.
• Proof Step 2: Multiply all parts of the inequality by $x^2$ (which is always positive).
• Proof Step 3: $-x^2 \leq x^2 \sin(\frac{1}{x}) \leq x^2$.
• Proof Step 4: Take the limit as $x \to 0$ of the outer bounds: $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$.
• Final Answer: By the Squeeze Theorem, the middle function's limit must also be $0$.

Problem 1.5: Limits Involving Absolute Values (Beginner)
Evaluate: $\lim_{x \to 2^-} \frac{|x-2|}{x-2}$

• Proof Step 1: Since $x$ approaches 2 from the left ($x \to 2^-$), $x$ is slightly less than 2.
• Proof Step 2: Therefore, $x - 2$ is negative.
• Proof Step 3: By definition of absolute value, if a quantity is negative, its absolute value is its negation: $|x-2| = -(x-2)$.
• Proof Step 4: Substitute this into the limit: $\lim_{x \to 2^-} \frac{-(x-2)}{x-2}$.
• Final Answer: The $(x-2)$ terms cancel, leaving exactly $-1$.

Problem 1.6: Limits with Radicals (Conjugates) (Intermediate)
Evaluate: $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

• Proof Step 1: Direct substitution yields $\frac{\sqrt{4} - 2}{4 - 4} = \frac{0}{0}$. Indeterminate form.
• Proof Step 2: Multiply numerator and denominator by the conjugate of the numerator: $\sqrt{x} + 2$.
• Proof Step 3: Numerator becomes $(\sqrt{x}-2)(\sqrt{x}+2) = x - 4$.
• Proof Step 4: Denominator is $(x-4)(\sqrt{x}+2)$.
• Proof Step 5: The limit is $\lim_{x \to 4} \frac{x-4}{(x-4)(\sqrt{x}+2)}$. Cancel the $(x-4)$ terms.
• Final Answer: Substitute $x=4$ into $\frac{1}{\sqrt{x}+2}$ to get $\frac{1}{2+2} = \frac{1}{4}$.

Category 2: Differentiation & Rates of Change

Problem 2.1: The Quotient Rule (Beginner)
Differentiate: $f(x) = \frac{3x - 1}{x^2 + 4}$

• Proof Step 1: Identify $u = 3x - 1$ and $v = x^2 + 4$.
• Proof Step 2: Compute derivatives: $u' = 3$ and $v' = 2x$.
• Proof Step 3: Apply formula $\frac{u'v - uv'}{v^2}$.
• Proof Step 4: $\frac{3(x^2 + 4) - (3x - 1)(2x)}{(x^2 + 4)^2}$.
• Proof Step 5: Distribute numerator: $3x^2 + 12 - (6x^2 - 2x) = -3x^2 + 2x + 12$.
• Final Answer: $f'(x) = \frac{-3x^2 + 2x + 12}{(x^2 + 4)^2}$.

Problem 2.2: Nesting The Chain Rule (Intermediate)
Differentiate: $y = \sin^3(4x)$

• Proof Step 1: Rewrite for clarity: $y = (\sin(4x))^3$.
• Proof Step 2: Outside function is $u^3$. Derivative is $3u^2 \cdot u'$. So, $3(\sin(4x))^2 \cdot \frac{d}{dx}[\sin(4x)]$.
• Proof Step 3: Now differentiate $\sin(4x)$. Outside is $\sin(v)$, inside is $4x$. Derivative is $\cos(4x) \cdot 4$.
• Proof Step 4: Multiply everything together: $3\sin^2(4x) \cdot 4\cos(4x)$.
• Final Answer: $y' = 12\sin^2(4x)\cos(4x)$.

Problem 2.3: Implicit Differentiation (Advanced)
Find $\frac{dy}{dx}$ for: $x^2 + y^2 = 2xy + 5$

• Proof Step 1: Differentiate both sides with respect to $x$. Remember to use the chain rule for $y$ (attach $\frac{dy}{dx}$).
• Proof Step 2: Left side derivative: $2x + 2y\frac{dy}{dx}$.
• Proof Step 3: Right side derivative (requires Product Rule for $2xy$): $2y + 2x\frac{dy}{dx}$. The $5$ becomes $0$.
• Proof Step 4: Equation is now $2x + 2y\frac{dy}{dx} = 2y + 2x\frac{dy}{dx}$.
• Proof Step 5: Move all $\frac{dy}{dx}$ terms to the left: $2y\frac{dy}{dx} - 2x\frac{dy}{dx} = 2y - 2x$.
• Proof Step 6: Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y - 2x) = 2y - 2x$.
• Final Answer: Divide both sides. $\frac{dy}{dx} = \frac{2y - 2x}{2y - 2x} = 1$. (Except where $y = x$).

Problem 2.4: Linear Approximation (Beginner)
Approximate: $\sqrt{16.1}$ using a tangent line.

• Proof Step 1: Choose a function $f(x) = \sqrt{x}$ and a "nice" base point $a = 16$.
• Proof Step 2: Evaluate the function and its derivative at $a$. $f(16) = 4$. $f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(16) = \frac{1}{8}$.
• Proof Step 3: Construct the tangent line equation: $L(x) = f(a) + f'(a)(x-a) = 4 + \frac{1}{8}(x-16)$.
• Proof Step 4: Plug in $x = 16.1$: $L(16.1) = 4 + \frac{1}{8}(16.1 - 16)$.
• Final Answer: $4 + \frac{1}{8}(0.1) = 4 + 0.0125 = 4.0125$.

Problem 2.5: Logarithmic Differentiation (Advanced)
Differentiate: $y = x^x$

• Proof Step 1: You cannot use the power rule because the exponent is a variable. Take the natural log of both sides: $\ln(y) = \ln(x^x)$.
• Proof Step 2: Use log properties to bring down the exponent: $\ln(y) = x\ln(x)$.
• Proof Step 3: Differentiate both sides implicitly. The left side is $\frac{1}{y} \cdot y'$. The right side requires the Product Rule: $1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.
• Proof Step 4: Multiply $y$ back over: $y' = y(\ln(x) + 1)$.
• Final Answer: Substitute $y = x^x$ back in: $y' = x^x(\ln(x) + 1)$.

Problem 2.6: Optimization (Maximizing Area) (Intermediate)
Problem: A farmer has 1000m of fencing to enclose a rectangular field bordered by a straight river (no fencing needed along the river). What dimensions maximize the area?

• Proof Step 1: Let the sides perpendicular to the river be $x$, and the side parallel be $y$. Perimeter constraint: $2x + y = 1000$.
• Proof Step 2: Area to maximize: $A = xy$.
• Proof Step 3: Solve constraint for $y$: $y = 1000 - 2x$. Substitute into Area: $A(x) = x(1000 - 2x) = 1000x - 2x^2$.
• Proof Step 4: Take derivative and set to zero to find critical points: $A'(x) = 1000 - 4x = 0$.
• Proof Step 5: Solving gives $x = 250$. Since $A''(x) = -4$ (concave down), this is a maximum.
• Final Answer: The dimensions are $x = 250$m and $y = 1000 - 2(250) = 500$m.

Category 3: Integration & Antiderivatives

Problem 3.1: Fundamental U-Substitution (Beginner)
Evaluate: $\int x^2 \sqrt{x^3 + 1} \, dx$

• Proof Step 1: Let $u = x^3 + 1$ (the inside of the radical).
• Proof Step 2: Differentiate to find $du = 3x^2 \, dx$. We need $x^2 \, dx$, so $\frac{1}{3}du = x^2 \, dx$.
• Proof Step 3: Substitute into the integral: $\int \sqrt{u} \cdot \frac{1}{3}du = \frac{1}{3} \int u^{1/2} \, du$.
• Proof Step 4: Use power rule: $\frac{1}{3} \cdot \frac{u^{3/2}}{3/2} = \frac{2}{9}u^{3/2}$.
• Final Answer: Back substitute $x$: $\frac{2}{9}(x^3 + 1)^{3/2} + C$.

Problem 3.2: Definite Integration by Parts (Intermediate)
Evaluate: $\int_0^{\pi} x \sin(x) \, dx$

• Proof Step 1: Use LIATE to choose $u$ and $dv$. $u = x$ (algebraic) and $dv = \sin(x) \, dx$ (trigonometric).
• Proof Step 2: Compute $du = 1 \, dx$ and $v = -\cos(x)$.
• Proof Step 3: Apply formula $uv - \int v \, du$: $-x\cos(x) - \int -\cos(x) \, dx$.
• Proof Step 4: Simplify the integral: $-x\cos(x) + \sin(x)$. This is the antiderivative.
• Proof Step 5: Apply bounds from $0$ to $\pi$. Evaluate at $\pi$: $-\pi\cos(\pi) + \sin(\pi) = -\pi(-1) + 0 = \pi$.
• Proof Step 6: Evaluate at $0$: $-0\cos(0) + \sin(0) = 0 + 0 = 0$.
• Final Answer: $\pi - 0 = \pi$.

Problem 3.3: Partial Fraction Decomposition (Advanced)
Evaluate: $\int \frac{3x+11}{x^2-x-6} \, dx$

• Proof Step 1: Factor the denominator: $(x-3)(x+2)$.
• Proof Step 2: Set up partial fractions: $\frac{A}{x-3} + \frac{B}{x+2} = \frac{3x+11}{(x-3)(x+2)}$.
• Proof Step 3: Multiply by denominator: $A(x+2) + B(x-3) = 3x+11$.
• Proof Step 4: Solve for constants. Let $x=3$: $5A = 20 \implies A = 4$. Let $x=-2$: $-5B = 5 \implies B = -1$.
• Proof Step 5: Rewrite integral: $\int \frac{4}{x-3} \, dx - \int \frac{1}{x+2} \, dx$.
• Proof Step 6: Integrate using the natural log rule.
• Final Answer: $4\ln|x-3| - \ln|x+2| + C$.

Problem 3.4: Area Between Two Curves (Beginner)
Find the area bounded by: $y = x^2$ and $y = \sqrt{x}$

• Proof Step 1: Find intersection points by setting them equal: $x^2 = \sqrt{x} \implies x^4 = x \implies x(x^3 - 1) = 0$. Intersections are at $x=0$ and $x=1$.
• Proof Step 2: On $[0,1]$, $\sqrt{x} \geq x^2$. So the area integral is $\int_0^1 (\sqrt{x} - x^2) \, dx$.
• Proof Step 3: Find antiderivative: $\frac{2}{3}x^{3/2} - \frac{1}{3}x^3$.
• Proof Step 4: Evaluate from 0 to 1: $(\frac{2}{3} - \frac{1}{3}) - (0 - 0)$.
• Final Answer: The bounded area is $\frac{1}{3}$.

Problem 3.5: Improper Integrals (Intermediate)
Evaluate: $\int_1^{\infty} \frac{1}{x^2} \, dx$

• Proof Step 1: Replace the infinite bound with a limit: $\lim_{t \to \infty} \int_1^t x^{-2} \, dx$.
• Proof Step 2: Find the antiderivative: $[-x^{-1}]_1^t = [-\frac{1}{x}]_1^t$.
• Proof Step 3: Evaluate at the bounds: $(-\frac{1}{t}) - (-\frac{1}{1}) = 1 - \frac{1}{t}$.
• Proof Step 4: Take the limit as $t \to \infty$. Since $\frac{1}{t} \to 0$, the limit evaluates cleanly.
• Final Answer: The integral converges to $1$.

Problem 3.6: Volumes of Solids of Revolution (Advanced)
Find the volume: The region bounded by $y=e^x$, $x=0$, $x=1$, and $y=0$ is rotated about the x-axis.

• Proof Step 1: Use the Disk Method. The volume formula is $V = \pi \int_a^b [R(x)]^2 \, dx$.
• Proof Step 2: The radius $R(x)$ is the function $e^x$. The bounds are $0$ to $1$.
• Proof Step 3: Set up the integral: $V = \pi \int_0^1 (e^x)^2 \, dx = \pi \int_0^1 e^{2x} \, dx$.
• Proof Step 4: The antiderivative of $e^{2x}$ is $\frac{1}{2}e^{2x}$.
• Proof Step 5: Evaluate from $0$ to $1$: $\pi [\frac{1}{2}e^{2(1)} - \frac{1}{2}e^{2(0)}]$.
• Final Answer: $\frac{\pi}{2}(e^2 - 1)$.

Category 4: Differential Equations

Problem 4.1: Separable Differential Equations (Intermediate)
Solve the Initial Value Problem: $\frac{dy}{dx} = x^2 y$, with $y(0) = 2$

• Proof Step 1: Separate the variables by moving $y$ to the left and $dx$ to the right: $\frac{1}{y} \, dy = x^2 \, dx$.
• Proof Step 2: Integrate both sides: $\int \frac{1}{y} \, dy = \int x^2 \, dx$.
• Proof Step 3: $\ln|y| = \frac{x^3}{3} + C$.
• Proof Step 4: Exponentiate both sides: $y = e^{\frac{x^3}{3} + C} = Ce^{\frac{x^3}{3}}$.
• Proof Step 5: Apply the initial condition $y(0) = 2$. $2 = Ce^0 \implies C = 2$.
• Final Answer: The particular solution is $y = 2e^{x^3/3}$.

Problem 4.2: First-Order Linear Equations (Advanced)
Solve: $y' + 2xy = x$

• Proof Step 1: Identify the integrating factor $I(x) = e^{\int 2x \, dx} = e^{x^2}$.
• Proof Step 2: Multiply the entire DE by the integrating factor: $e^{x^2}y' + 2xe^{x^2}y = xe^{x^2}$.
• Proof Step 3: The left side collapses perfectly into the derivative of a product: $\frac{d}{dx}[ye^{x^2}] = xe^{x^2}$.
• Proof Step 4: Integrate both sides: $ye^{x^2} = \int xe^{x^2} \, dx$. Use u-substitution on the right ($u=x^2$).
• Proof Step 5: $ye^{x^2} = \frac{1}{2}e^{x^2} + C$.
• Final Answer: Isolate $y$: $y = \frac{1}{2} + Ce^{-x^2}$.

Problem 4.3: Newton's Law of Cooling (Intermediate)
Problem: A $90^\circ C$ cup of coffee sits in a $20^\circ C$ room. After 10 mins, it's $60^\circ C$. When will it reach $35^\circ C$?

• Proof Step 1: The model is $T(t) = T_s + (T_0 - T_s)e^{kt}$. Substitute knowns: $T(t) = 20 + 70e^{kt}$.
• Proof Step 2: Use the 10-minute condition to find $k$: $60 = 20 + 70e^{10k} \implies 40 = 70e^{10k} \implies e^{10k} = \frac{4}{7}$.
• Proof Step 3: $10k = \ln(\frac{4}{7}) \implies k \approx -0.05596$.
• Proof Step 4: Now solve for $t$ when $T = 35$: $35 = 20 + 70e^{-0.05596t}$.
• Proof Step 5: $15 = 70e^{-0.05596t} \implies \frac{15}{70} = e^{-0.05596t}$.
• Final Answer: $t = \frac{\ln(3/14)}{-0.05596} \approx 27.5$ minutes.

Category 5: Infinite Sequences & Series

Problem 5.1: The Ratio Test (Intermediate)
Determine convergence for: $\sum_{n=1}^{\infty} \frac{2^n}{n!}$

• Proof Step 1: Apply the Ratio Test: $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$.
• Proof Step 2: $a_{n+1} = \frac{2^{n+1}}{(n+1)!}$.
• Proof Step 3: Divide: $\frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n}$.
• Proof Step 4: Simplify. $2^{n+1}/2^n = 2$, and $n!/(n+1)! = \frac{1}{n+1}$.
• Proof Step 5: The limit is $\lim_{n \to \infty} \frac{2}{n+1} = 0$.
• Final Answer: Since $0 < 1$, the series converges absolutely.

Problem 5.2: Maclaurin Series Derivation (Advanced)
Find the first 4 non-zero terms of the Maclaurin series for: $f(x) = \sin(x)$

• Proof Step 1: Evaluate $f(x)$ and its derivatives at $x=0$.
• Proof Step 2: $f(0) = \sin(0) = 0$. $f'(0) = \cos(0) = 1$. $f''(0) = -\sin(0) = 0$. $f'''(0) = -\cos(0) = -1$. $f^{(4)}(0) = \sin(0) = 0$. $f^{(5)}(0) = \cos(0) = 1$.
• Proof Step 3: Plug into Taylor formula: $\sum \frac{f^{(n)}(0)}{n!} x^n$.
• Proof Step 4: Terms are: $\frac{0}{0!}x^0 + \frac{1}{1!}x^1 + \frac{0}{2!}x^2 + \frac{-1}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5$.
• Final Answer: $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

Problem 5.3: Finding the Radius of Convergence (Advanced)
Find the interval of convergence for: $\sum_{n=1}^{\infty} \frac{(x-3)^n}{n}$

• Proof Step 1: Use the Ratio Test: $\lim_{n \to \infty} |\frac{(x-3)^{n+1}}{n+1} \cdot \frac{n}{(x-3)^n}|$.
• Proof Step 2: This simplifies to $\lim_{n \to \infty} |\frac{n}{n+1} (x-3)| = |x-3|$.
• Proof Step 3: For convergence, we need $|x-3| < 1$, which means $-1 < x-3 < 1 \implies 2 < x < 4$. The radius is $R=1$.
• Proof Step 4: Check endpoints. If $x=4$, series is $\sum \frac{1}{n}$ (harmonic, diverges). If $x=2$, series is $\sum \frac{(-1)^n}{n}$ (alternating harmonic, converges).
• Final Answer: The interval of convergence is $[2, 4)$.