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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 2 - LimitsProblem 2.7.31
Chapter 2, Problem 2.7.31

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.
lim x→−3 |2x|=6 (Hint: Use the inequality ∥a|−|b∥≤|a−b|, which holds for all constants a and b (see Exercise 74).)

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Step 1: Start by recalling the precise definition of a limit. For the limit \( \lim_{x \to -3} |2x| = 6 \) to hold, for every \( \varepsilon > 0 \), there must exist a \( \delta > 0 \) such that if \( 0 < |x + 3| < \delta \), then \( ||2x| - 6| < \varepsilon \).
Step 2: Express the function in terms of the limit. We have \( |2x| = 2|x| \). Therefore, we need to show that \( |2|x| - 6| < \varepsilon \) whenever \( 0 < |x + 3| < \delta \).
Step 3: Simplify the inequality \( |2|x| - 6| < \varepsilon \). This can be rewritten as \( |2x - 6| < \varepsilon \).
Step 4: Relate \( |2x - 6| \) to \( |x + 3| \). Notice that \( |2x - 6| = 2|x + 3| \). Therefore, the inequality \( |2x - 6| < \varepsilon \) becomes \( 2|x + 3| < \varepsilon \).
Step 5: Solve for \( \delta \) in terms of \( \varepsilon \). From \( 2|x + 3| < \varepsilon \), we get \( |x + 3| < \frac{\varepsilon}{2} \). Thus, choose \( \delta = \frac{\varepsilon}{2} \) to satisfy the condition.

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Key Concepts

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

Limit Definition

The precise definition of a limit states that for a function f(x) to approach a limit L as x approaches a value c, for every ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε. This formalizes the intuitive idea of limits and is essential for proving limit statements rigorously.
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Absolute Value and Inequalities

The absolute value function measures the distance of a number from zero on the number line. The inequality ∥a|−|b∥≤|a−b| helps in establishing relationships between the absolute values of expressions, which is crucial when manipulating limits involving absolute values, such as |2x| in this case.
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Epsilon-Delta Relationship

In limit proofs, the relationship between ε and δ is critical. It specifies how close x must be to c (within δ) to ensure that f(x) is within ε of L. Establishing this relationship is key to demonstrating that the limit exists and is valid under the conditions set by the definition of a limit.
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Related Practice
Textbook Question

Use the precise definition of infinite limits to prove the following limits.


limx→−11(x+1)4=∞{\(\displaystyle\]\lim\)_{x\(\to\)-1}}\(\frac{1}{\left(x+1\right)^4}\)=\(\infty\)

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Textbook Question

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.

lim x→1 (8x+5)=13

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Textbook Question

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.

lim x→5 1/x^2=1/25

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Textbook Question

Which of the following statements are correct? Choose all that apply.


a. lim x→1 1/ (x−1)^2 does not exist


b. lim x→1 1/ (x−1)^2=∞


c. lim x→1 1/(x−1)^2=−∞

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Textbook Question

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.

lim x→4 x^2−16 / x−4=8 (Hint: Factor and simplify.)

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Textbook Question

Determine the following limits at infinity.


lim x→−∞ 3x^11

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