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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 2.7.39

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→a (mx+b)=ma+b, for any constants a, b, and m

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Step 1: Start by recalling the precise definition of a limit. For a function f(x), we say that \( \lim_{x \to a} f(x) = L \) if for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - a| < \delta \), it follows that \( |f(x) - L| < \varepsilon \).
Step 2: In this problem, the function is \( f(x) = mx + b \) and the limit we want to prove is \( \lim_{x \to a} (mx + b) = ma + b \). Therefore, we need to show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |(mx + b) - (ma + b)| < \varepsilon \).
Step 3: Simplify the expression \( |(mx + b) - (ma + b)| \). This simplifies to \( |mx + b - ma - b| = |mx - ma| = |m||x - a| \).
Step 4: To ensure \( |m||x - a| < \varepsilon \), we can choose \( \delta \) such that \( |x - a| < \frac{\varepsilon}{|m|} \). Therefore, set \( \delta = \frac{\varepsilon}{|m|} \) if \( m \neq 0 \). If \( m = 0 \), the function is constant, and the limit is trivially \( b \), so any \( \delta > 0 \) will work.
Step 5: Conclude that by choosing \( \delta = \frac{\varepsilon}{|m|} \) (or any positive \( \delta \) if \( m = 0 \)), we have shown that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that \( 0 < |x - a| < \delta \) implies \( |(mx + b) - (ma + b)| < \varepsilon \), thus proving the limit.

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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 a, for every ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε. This formalism is crucial for proving limits rigorously.
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Linear Functions

The expression mx + b represents a linear function, where m is the slope and b is the y-intercept. Understanding the behavior of linear functions as x approaches a specific value is essential for evaluating limits, as they exhibit predictable and continuous behavior.
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Epsilon-Delta Relationship

In limit proofs, establishing a relationship between ε and δ is vital. For the limit lim x→a (mx + b) = ma + b, one can show that choosing δ = ε/|m| ensures that the condition |f(x) - L| < ε is satisfied, thereby proving the limit exists.
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