Margaret L. Lial • Raymond N. Greenwell • Nathan P. Ritchey
ISBN #9781292108971
11th Edition
3,612 Questions
Homework Questions
This section introduces the fundamental concept of a limit, which underlies all of calculus. It explains how limits allow us to analyze function behavior near specific points even when the function might be undefined at that point. The text demonstrates various methods to evaluate limits—algebraically, by using tables, and with graphing calculators—and highlights the importance of one-sided limits and handling indeterminate forms. Limits at infinity are also discussed, linking them to horizontal asymptotes and long-term behavior of functions.
1
Describe the concept of a limit and explain its role in understanding rates of change in dynamic situations.
2
Identify and compute one-sided limits and two-sided limits using tables, graphs, and algebraic methods.
3
Apply limit rules—including handling indeterminate forms—to evaluate limits of functions.
4
Evaluate limits at infinity to determine horizontal asymptotes and understand long?term function behavior.
CONCEPT
DEFINITION
Limit
The value that a function f(x) approaches as x gets arbitrarily close to a specific value a (without necessarily equaling f(a)).
One-Sided Limit
The value that f(x) approaches as x approaches a from one side only (written as x → a⁻ for the left-hand limit, and x → a⁺ for the right-hand limit).
Indeterminate Form
An expression like 0/0 that does not have a clear meaning until it is simplified, usually by factoring or other algebraic techniques.
Limit at Infinity
The behavior of f(x) as x becomes arbitrarily large (x → ∞) or arbitrarily small (x → −∞), often indicating a horizontal asymptote.
Horizontal Asymptote
A horizontal line y = L that the graph of a function approaches as x → ∞ or x → −∞, corresponding to lim x→∞ f(x) = L (or lim x→ −∞ f(x) = L).
In Exercises 1-4, choose the best answer for each limit. $$If \lim _{x \rightarrow 2} f(x)=5 and \lim _{x \rightarrow 2^{-}} f(x)=6,$ then \lim _{x \rightarrow 2} f(x)$$ $$\begin{array}{ll}{\text { (a) is } 5 .} & {\text { (b) is } 6} \\ {\text { (c) does not exist. }} & {\text { (d) is infinite. }}\end{array}$$
In Exercises 1-4, choose the best answer for each limit. $$If \lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=-1, but f(2)=1, then \lim _{x \rightarrow 2} f(x)$$ $$\begin{array}{ll}{\text { (a) is }-1 .} & {\text { (b) does not exist. }} \\ {\text { (c) is infinite. }} & {\text { (d) is } 1 .}\end{array}$$
In Exercises 1-4, choose the best answer for each limit. $$If \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=5, but f(4) does not exist, then \lim _{x \rightarrow 4^{-}} f(x)$$ $$\begin{array}{ll}{\text { (a) is } 5} & {\text { (b) is }-\infty} \\ {\text { (c) is }+\infty} & {\text { (d) does not exist }}\end{array}$$
In Exercises 1-4, choose the best answer for each limit. $$If \lim _{x \rightarrow 1} f(x)=-\infty and \lim _{x \rightarrow 1^{+}} f(x)=-\infty, then \lim _{x \rightarrow 1} f(x)$$ $$\begin{array}{ll}{\text { (a) is } \infty .} & {\text { (b) is }-\infty.} \\ {\text { (c) does not exist. }} & {\text { (d) is } 1 \text { . }}\end{array}$$
Decide whether each limit exists. If a limit exists, estimate its value. $${ (a) }\lim _{x \rightarrow 3} f(x) \quad \text { (b) } \lim _{x \rightarrow 0} f(x) $$
QUESTION
What is lim x→2 f(x) where f(x) = x²?
STEP-BY-STEP ANSWER:
Step 1: Create a table of values where x is close to 2 (for example: 1.9, 1.99, 2.01, 2.1). Step 2: Compute f(x) = x² for each x value. Step 3: Observe that as x approaches 2, the values approach 4. Final Answer: lim x→2 x² = 4.
Evaluating a Limit via Table (f(x) = x² as x → 2)
Find lim x→2 g(x) given that g(x) = (x³ - 2x²)/(x - 2) is undefined at x = 2 but can be simplified.
Step 1: Factor the numerator: x³ - 2x² = x²(x - 2). Step 2: Simplify the expression: g(x) = x²(x - 2)/(x - 2) = x² for all x ≠ 2. Step 3: Evaluate the limit: lim x→2 x² = 2² = 4. Final Answer: lim x→2 g(x) = 4.
Evaluating a One-Sided Limit (g(x) = (x³ - 2x²)/(x - 2) as x → 2)
Determine lim x→2 h(x) where h(x) = (x² + x - 6)/(x - 2) results in 0/0 when directly substituted.
Step 1: Factor the numerator: x² + x - 6 = (x + 3)(x - 2). Step 2: Cancel the common factor (x - 2) from numerator and denominator, yielding h(x) = x + 3 for x ≠ 2. Step 3: Evaluate the limit: lim x→2 (x + 3) = 2 + 3 = 5. Final Answer: lim x→2 h(x) = 5.
Handling an Indeterminate Form (h(x) = (x² + x - 6)/(x - 2) as x → 2)