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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 2.7.25
Chapter 2, Problem 2.7.25

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→7 f(x)=9, where f(x)={3x−12 if x≤7
x+2 if x>7

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We need to prove that \( \lim_{{x \to 7}} f(x) = 9 \) using the precise definition of a limit. The function \( f(x) \) is defined piecewise: \( f(x) = 3x - 12 \) for \( x \leq 7 \) and \( f(x) = x + 2 \) for \( x > 7 \).
The precise definition of a limit states that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 7| < \delta \), then \( |f(x) - 9| < \varepsilon \).
For \( x \leq 7 \), \( f(x) = 3x - 12 \). We need \( |(3x - 12) - 9| < \varepsilon \). Simplifying, this becomes \( |3x - 21| < \varepsilon \), or \( |3(x - 7)| < \varepsilon \).
From \( |3(x - 7)| < \varepsilon \), we have \( |x - 7| < \frac{\varepsilon}{3} \). Thus, for \( x \leq 7 \), we can choose \( \delta = \frac{\varepsilon}{3} \).
7 \).> For \( x > 7 \), \( f(x) = x + 2 \). We need \( |(x + 2) - 9| < \varepsilon \). Simplifying, this becomes \( |x - 7| < \varepsilon \). Thus, for \( x > 7 \), we can choose \( \delta = \varepsilon \).

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Key Concepts

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

Definition of a Limit

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 must exist a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε. This formal definition is crucial for proving limits rigorously.
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Piecewise Functions

A piecewise function is defined by different expressions based on the input value. In this case, f(x) has two definitions: 3x - 12 for x ≤ 7 and x + 2 for x > 7. Understanding how to evaluate limits at points where the function changes its definition is essential for solving the limit problem.
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Epsilon-Delta Proof

An epsilon-delta proof involves demonstrating that for any chosen ε, a corresponding δ can be found that satisfies the limit definition. This method is used to show that as x approaches a certain value, the function's output can be made arbitrarily close to the limit, thus confirming the limit's existence.
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Related Practice
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Let g(x)={x2+xif x<1aif x=13x+5if x>1g\(\left\)(x\(\right\))=\(\begin{cases}\)x^2+x & \(\text{if }\)x<1\\ a & \(\text{if }\)x=1\\ 3x+5 & \(\text{if }\)x>1\(\end{cases}\)

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