Welcome to the comprehensive practice module for Week 4, Lecture 10. Work through these 20 scaffolded questions covering punctured neighborhoods, indeterminate forms ($\frac{0}{0}$), factoring, radical conjugation, one-sided limits, the Rendezvous Criterion, limit laws, and the Squeeze Theorem. Click any option for instant feedback and detailed explanations.
Module 1: Punctured Neighborhoods & Formal Definitions
Question 1: The Punctured Neighborhood Constraint
What is the primary mathematical purpose of the inequality $0 < |x - c| < \delta$ in the formal definition of a limit, compared to the continuity condition $|x - c| < \delta$?
A) It forces the function to be continuous at $x = c$.
B) It creates a punctured neighborhood, strictly forbidding $x = c$ so we can evaluate limits at holes.
C) It ensures the limit value $L$ is always equal to $f(c)$.
D) It guarantees that the denominator can never equal zero anywhere on the real line.
💡 Toggle Hint
Think about what happens when $x = c$ causes $\frac{0}{0}$. Why do we want to exclude $x = c$ from our test window?
Question 2: Ignoring Displaced Points
Suppose a function $f(x)$ has an open hole at $x = 2$ where the limit trajectory aims at $L = 5$, but the actual function value is defined separately as $f(2) = 10$. What does $\lim_{x \to 2} f(x)$ evaluate to?
A) $10$, because function values always override limits.
B) $5$, because limits use a punctured neighborhood and completely ignore what happens at $x = c$.
C) The limit Does Not Exist (DNE) because $f(2) \neq L$.
D) $7.5$, the average of the hole and the displaced point.
💡 Toggle Hint
Review the geometric anatomy diagram in Section 1. Does a limit care about where $f(c)$ is plotted?
Question 3: Adversary vs. Defender Roles
In the $\epsilon\text{-}\delta$ game defining limits, what is the role of the Adversary who chooses $\epsilon > 0$?
A) To calculate the exact numerical limit $L$.
B) To challenge your proof by handing you an arbitrarily tight output error corridor $\epsilon$.
C) To forbid $x$ from ever approaching $c$.
D) To factor polynomials and rationalize radicals.
💡 Toggle Hint
Recall Player 1 from our game mechanics in Lecture 9 and Lecture 10.
Question 4: Domain Requirements for Limits
For a two-sided limit $\lim_{x \to c} f(x)$ to be meaningfully evaluated, what must be true about the domain of $f(x)$?
A) The function must be continuous across the entire real number line.
B) The function must be defined on an open interval containing $c$, except possibly at $c$ itself.
C) The function must be a polynomial of degree 2 or higher.
D) $f(c)$ must equal zero.
💡 Toggle Hint
Check the formal definition box in Section 1. What interval requirement is stated right at the beginning?
Module 2: Algebraic Surgery & Factoring ($\frac{0}{0}$)
Question 5: Basic Polynomial Factoring Limit
Evaluate the limit: $$\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$$
A) $0$
B) $4$
C) $8$
D) Does Not Exist (DNE)
💡 Toggle Hint
Factor the numerator as a difference of squares: $x^2 - 16 = (x - 4)(x + 4)$. Then cancel $(x - 4)$.
Question 6: Quadratic Factoring
Evaluate the limit: $$\lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2}$$
A) $1$
B) $-1$
C) $2$
D) $0$
💡 Toggle Hint
Factor the numerator $x^2 + 5x + 6$ into $(x + 2)(x + 3)$.
Question 7: Difference of Cubes
Evaluate the limit: $$\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$$
A) $4$
B) $6$
C) $12$
D) $0$
💡 Toggle Hint
Recall the difference of cubes formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Here $a = x, b = 2$.
Question 8: Non-Zero over Zero Asymptote
Evaluate the limit: $$\lim_{x \to 3^+} \frac{x + 5}{x - 3}$$
A) $0$
B) $8$
C) $+\infty$
D) $-\infty$
💡 Toggle Hint
Direct substitution gives $\frac{8}{0^+}$. A positive number divided by an infinitesimally small positive number shoots to positive infinity!
Module 3: Radical Conjugation & Rationalization
Question 9: Basic Radical Conjugate
Evaluate the limit: $$\lim_{x \to 0} \frac{\sqrt{x + 16} - 4}{x}$$
A) $0$
B) $\frac{1}{4}$
C) $\frac{1}{8}$
D) $8$
💡 Toggle Hint
Multiply numerator and denominator by the conjugate $\sqrt{x + 16} + 4$. The numerator simplifies to $(x + 16) - 16 = x$.
Question 10: Shifted Radical Conjugate
Evaluate the limit: $$\lim_{x \to 5} \frac{\sqrt{x - 1} - 2}{x - 5}$$
A) $\frac{1}{2}$
B) $\frac{1}{4}$
C) $2$
D) $0$
💡 Toggle Hint
Multiply top and bottom by $\sqrt{x - 1} + 2$. The numerator becomes $(x - 1) - 4 = x - 5$, which cancels $(x - 5)$ in the denominator.
Question 11: Radical in Denominator
Evaluate the limit: $$\lim_{x \to 0} \frac{x}{\sqrt{x + 4} - 2}$$
A) $2$
B) $\frac{1}{4}$
C) $4$
D) $0$
💡 Toggle Hint
Multiply numerator and denominator by the conjugate of the denominator: $\sqrt{x + 4} + 2$.
Question 12: Complex Conjugate Form
Evaluate the limit: $$\lim_{x \to 1} \frac{\sqrt{8 + x} - 3}{x - 1}$$
A) $\frac{1}{3}$
B) $\frac{1}{6}$
C) $6$
D) $3$
💡 Toggle Hint
Conjugate is $\sqrt{8 + x} + 3$. Numerator simplifies to $(8 + x) - 9 = x - 1$.
Module 4: One-Sided Limits & The Rendezvous Criterion
Question 13: Absolute Value Left-Hand Limit
Evaluate the left-hand limit: $$\lim_{x \to 0^-} \frac{|x|}{x}$$
A) $1$
B) $-1$
C) $0$
D) Does Not Exist (DNE)
💡 Toggle Hint
For $x < 0$, $|x| = -x$. Thus $\frac{-x}{x} = -1$.
Question 14: Rendezvous Criterion Failure
Given a piecewise function where $\lim_{x \to c^-} f(x) = 5$ and $\lim_{x \to c^+} f(x) = 8$, what can we conclude about $\lim_{x \to c} f(x)$?
A) The limit is $6.5$.
B) The limit is $5$.
C) The two-sided limit Does Not Exist (DNE) because the one-sided limits disagree.
D) The limit is $8$.
💡 Toggle Hint
Check the Rendezvous Criterion in Section 3. Do the bridge crews meet?
Question 15: Piecewise Function Limit
Let $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 3x - 1 & \text{if } x \ge 2 \end{cases}$. What is $\lim_{x \to 2^-} f(x)$?
A) $3$
B) $5$
C) $2$
D) $4$
💡 Toggle Hint
For $x < 2$, use the first sub-function $x^2 + 1$ and substitute $x = 2$.
Question 16: Two-Sided Piecewise Check
Using the same piecewise function from Question 15 ($f(x) = x^2+1$ for $x < 2$ and $3x-1$ for $x \ge 2$), does $\lim_{x \to 2} f(x)$ exist?
A) Yes, because both left-hand and right-hand limits equal $5$.
B) No, because there is a jump discontinuity.
C) No, because $f(2)$ is undefined.
D) Yes, with a limit of $2$.
💡 Toggle Hint
Calculate both $\lim_{x \to 2^-} (x^2+1) = 5$ and $\lim_{x \to 2^+} (3x-1) = 5$. Compare them!
Module 5: Limit Laws & The Squeeze Theorem
Question 17: Squeeze Theorem Application
If $3x \le f(x) \le x^3 + 2x$ for all $x$, what is $\lim_{x \to 1} f(x)$?
A) $3$
B) $1$
C) $2$
D) Limit Does Not Exist
💡 Toggle Hint
Evaluate the limit of the floor $3(1) = 3$ and ceiling $(1)^3 + 2(1) = 3$ as $x \to 1$.
Question 18: Landmark Trigonometric Limit
Evaluate the fundamental limit: $$\lim_{x \to 0} \frac{\sin(x)}{x}$$
A) $0$
B) $1$
C) $\infty$
D) Undefined
💡 Toggle Hint
Proved via the Squeeze Theorem in Section 5 (trapped between $\cos(x)$ and $1$).
Question 19: Scaled Trig Limit
Evaluate the limit: $$\lim_{x \to 0} \frac{\sin(5x)}{x}$$
A) $1$
B) $\frac{1}{5}$
C) $5$
D) $0$
💡 Toggle Hint
Multiply top and bottom by $5$: $\frac{5\sin(5x)}{5x} = 5 \cdot \frac{\sin(5x)}{5x}$. As $5x \to 0$, this evaluates to $5 \cdot 1 = 5$.
Question 20: Limit Law Quotient Failure
If $\lim_{x \to c} f(x) = 4$ and $\lim_{x \to c} g(x) = 0$, what can we immediately state about $\lim_{x \to c} \frac{f(x)}{g(x)}$?
A) The limit is $0$ because numerator is $4$.
B) The limit is $\infty$ always.
C) The Quotient Law cannot be applied directly because the denominator limit is zero (producing a $\frac{4}{0}$ asymptote or indeterminate form).
D) The limit is $1$.
💡 Toggle Hint
Review Section 4 on the Quotient Law warning label ($M \neq 0$).