Week 4, Lecture 11 Practice Problems: Limits at Infinity

Welcome to the comprehensive practice module for Week 4, Lecture 11. Work through these 10 scaffolded questions covering limits at infinity, horizontal and vertical asymptotes, and the dominant power race for rational functions.

Module 1: Horizontal Asymptotes & The Dominant Power Race

Question 1: Bottom Heavy (Denominator Wins)

Evaluate the limit: $$\lim_{x \to \infty} \frac{5x^2 + 2x - 1}{x^3 + 4x}$$

The highest power of $x$ is in the denominator ($x^3$). This makes the rational function bottom-heavy.

Question 2: Equal Powers (A Dead Tie)

Find the horizontal asymptote of the function: $$f(x) = \frac{8x^4 - 3x^2 + 1}{2x^4 + 5x^3}$$

The highest power of $x$ is $x^4$ in both the numerator and denominator. Divide the leading coefficients: $\frac{8}{2}$.

Question 3: Top Heavy (Numerator Wins)

Evaluate the limit: $$\lim_{x \to -\infty} \frac{x^3 + 2}{x^2 - 1}$$

The function is top-heavy ($x^3$ vs $x^2$), so it grows without bound. Notice that $x$ approaches negative infinity. A negative number cubed is negative, divided by a negative number squared (positive), resulting in a negative value.

Question 4: Trick Question (Mismatched Order)

Evaluate the limit: $$\lim_{x \to \infty} \frac{4 - 3x^3}{5x^3 + x - 2}$$

Don't just look at the first term! Identify the term with the highest power of $x$ in both the numerator ($-3x^3$) and the denominator ($5x^3$).

Question 5: Rigorous Algebraic Evaluation

To evaluate $\lim_{x \to \infty} \frac{2x^2 + 3x}{5x^2 - 1}$ rigorously without using the "leading coefficient" shortcut, you must divide every term by the highest power of $x$ in the denominator. What is the simplified form of the fraction before applying the limit as $x \to \infty$?

The highest power of $x$ in the denominator is $x^2$. Divide every single term ($2x^2, 3x, 5x^2, -1$) by $x^2$ and simplify.

Module 2: Vertical Asymptotes & Infinite Limits

Question 6: Vertical Asymptote Identification

Which of the following $x$-values represents a vertical asymptote for the function $f(x) = \frac{x - 3}{x^2 - 9}$?

Factor the denominator: $x^2 - 9 = (x - 3)(x + 3)$. The term $(x - 3)$ cancels with the numerator, leaving a hole at $x = 3$. The remaining term $(x + 3)$ in the denominator causes a $\frac{c}{0}$ explosion at $x = -3$.

Question 7: Infinite Limits from the Left

Evaluate the one-sided limit: $$\lim_{x \to 2^-} \frac{1}{x - 2}$$

Since $x$ is approaching $2$ from the left, $x$ is slightly smaller than $2$ (e.g., $1.99$). The denominator $x - 2$ will be a very small negative number. Dividing a positive constant by a tiny negative number yields $-\infty$.

Question 8: Infinite Limits from the Right

Evaluate the one-sided limit: $$\lim_{x \to 5^+} \frac{-4}{x - 5}$$

As $x \to 5^+$, the denominator $(x - 5)$ is a small positive number. The numerator is a negative constant ($-4$). A negative divided by a positive yields negative infinity.

Question 9: Limit Definition of $e$

Evaluate the limit: $$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$$

This is the formal definition of Euler's number, representing continuous compounding growth. See Section 4.

Question 10: Horizontal Asymptotes of Transcendental Functions

Evaluate the limit: $$\lim_{x \to \infty} e^{-x}$$

Recall that $e^{-x} = \frac{1}{e^x}$. As $x \to \infty$, the denominator $e^x$ grows infinitely large. Thus, the fraction approaches $0$.

Week 4, Lecture 11: Practice Problem Set (10 Questions)