Ch. 2 - Limits

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 2 - Limits

Problem 86

Chapter 2, Problem 86

Use an appropriate limit definition to prove the following limits.

lim x→1 (5x−2) =3;

Verified step by step guidance

Start by recalling the definition of a limit: \( \lim_{{x \to a}} f(x) = L \) means that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon \).

Identify the function \( f(x) = 5x - 2 \), the point \( a = 1 \), and the limit \( L = 3 \).

Set up the inequality \( |f(x) - L| < \epsilon \) which becomes \( |(5x - 2) - 3| < \epsilon \). Simplify this to \( |5x - 5| < \epsilon \).

Factor the expression inside the absolute value: \( |5(x - 1)| < \epsilon \). This simplifies to \( 5|x - 1| < \epsilon \).

Solve for \( |x - 1| \) by dividing both sides by 5: \( |x - 1| < \frac{\epsilon}{5} \). Choose \( \delta = \frac{\epsilon}{5} \) to satisfy the limit definition.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limit Definition

The limit definition in calculus refers to the formal approach to determining the value that a function approaches as the input approaches a certain point. Specifically, for a function f(x), the limit as x approaches a value 'a' is L if, for every ε > 0, there exists a δ > 0 such that whenever 0 < |x - a| < δ, it follows that |f(x) - L| < ε. This definition is foundational for proving limits rigorously.

Direct Substitution

Direct substitution is a method used to evaluate limits by substituting the value that x approaches directly into the function. If the function is continuous at that point, the limit can be found simply by replacing x with the target value. In the case of the limit lim x→1 (5x−2), substituting x = 1 yields the result 3, confirming the limit without further manipulation.

Continuity of Functions

A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. For the limit lim x→1 (5x−2), the function is a polynomial, which is continuous everywhere. This property allows us to confidently use direct substitution to evaluate the limit, reinforcing the relationship between limits and continuity in calculus.

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Related Practice

Textbook Question

Suppose f(x) = {x^2 − 5x + 6 / x − 3 if x≠3

a if x=3.

Determine a value of the constant a for which lim x→3 f(x) = f(3).

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Textbook Question

a. Use the Intermediate Value Theorem to show that the equation has a solution in the given interval.

x=cos x; (0,π/2)

Textbook Question

A sine limit It can be shown that 1−x^2/ 6 ≤ sin x/ x ≤1, for x near 0.

Use these inequalities to evaluate lim x→0 sin x/ x.

Textbook Question

Consider the graph of y=cot^−1 x(see Section 1.4) and determine the following limits using the graph.

lim x→−∞ cot^−1x

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Textbook Question

Calculate the following limits using the factorization formula x^n−a^n=(x−a)(x^n−1+ax^n−2+a^2x^n−3+⋯+a^n−2x+a^n−1), where n is a positive integer and a is a real number.

lim x→1 x^6 − 1 / x − 1

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Textbook Question

Suppose g(x) = {x^2−5x if x≤−1

ax^3−7 if x>−1.

Determine a value of the constant a for which lim x→−1 g(x) exists and state the value of the limit, if possible.

Use an appropriate limit definition to prove the following limits.lim x→1 (5x−2) =3;

Verified video answer for a similar problem:

Key Concepts

Limit Definition

Direct Substitution

Continuity of Functions

Use an appropriate limit definition to prove the following limits.

lim x→1 (5x−2) =3;