Math Problem Statement

Prove the following limit. lim x → 7 (3x − 17) = 4 SOLUTION Preliminary analysis of the problem (guessing a value for 𝛿). Let 𝜀 be a given positive number. We want to find a number 𝛿 such that if 0 < |x − 7| < 𝛿 then |(3x − 17) − 4| < 𝜀. But |(3x − 17) − 4| = |3x − 21| = 3

. Therefore, we want 𝛿 such that if 0 < |x − 7| < 𝛿 then 3

< 𝜀 that is, if 0 < |x − 7| < 𝛿 then

< 𝜀 3 . This suggests that we should choose 𝛿 = 𝜀 3 . Proof (showing that 𝛿 works). Given 𝜀 > 0, choose 𝛿 = 𝜀 3 . If 0 <

< 𝛿, then we get the following. |(3x − 17) − 4| =

= 3

< 3𝛿 = 3

𝜀

= 𝜀 Thus, if 0 < |x − 7| < 𝛿 then |(3x − 17) − 4| < 𝜀. Therefore, by the definition of a limit, we get the following. lim x → 7 (3x − 17) = 4

Solution

We are asked to prove the limit:

limx7(3x17)=4\lim_{x \to 7} (3x - 17) = 4

Preliminary analysis

We need to prove the limit using the formal ϵ\epsilon-δ\delta definition of a limit. Specifically, given any ϵ>0\epsilon > 0, we want to find a corresponding δ>0\delta > 0 such that if 0<x7<δ0 < |x - 7| < \delta, then:

(3x17)4<ϵ| (3x - 17) - 4 | < \epsilon

Step 1: Simplify the expression (3x17)4|(3x - 17) - 4|

We start by simplifying the expression inside the absolute value:

(3x17)4=3x21| (3x - 17) - 4 | = | 3x - 21 |

Factoring out a 3 from the absolute value:

3x21=3x7| 3x - 21 | = 3 |x - 7|

Step 2: Set the inequality 3(x7)<ϵ|3(x - 7)| < \epsilon

Now, for the limit to hold, we need:

3x7<ϵ3 |x - 7| < \epsilon

Dividing both sides of the inequality by 3:

x7<ϵ3|x - 7| < \frac{\epsilon}{3}

This suggests that we should choose δ=ϵ3\delta = \frac{\epsilon}{3}.

Proof

To formally prove the limit, we proceed as follows:

Let ϵ>0\epsilon > 0 be given. Choose δ=ϵ3\delta = \frac{\epsilon}{3}. Now, suppose 0<x7<δ0 < |x - 7| < \delta. Then:

x7<δ=ϵ3|x - 7| < \delta = \frac{\epsilon}{3}

Multiplying both sides by 3:

3x7<3×ϵ3=ϵ3 |x - 7| < 3 \times \frac{\epsilon}{3} = \epsilon

Thus, we have:

3x21=3x7<ϵ|3x - 21| = 3 |x - 7| < \epsilon

Therefore, if 0<x7<δ0 < |x - 7| < \delta, then (3x17)4=3x21<ϵ|(3x - 17) - 4| = |3x - 21| < \epsilon.

This confirms that the limit holds, and by the definition of a limit, we conclude:

limx7(3x17)=4\lim_{x \to 7} (3x - 17) = 4

Do you need any further details or have any questions?

Here are some related questions to expand on this topic:

  1. How does the ϵ\epsilon-δ\delta definition of limits apply to nonlinear functions?
  2. Can you explain the geometrical interpretation of the ϵ\epsilon-δ\delta definition of limits?
  3. How would the ϵ\epsilon-δ\delta proof change if the function were quadratic instead of linear?
  4. What is the difference between one-sided and two-sided limits in terms of ϵ\epsilon-δ\delta?
  5. How can you use ϵ\epsilon-δ\delta limits to prove the continuity of a function at a point?

Tip: For any linear function f(x)=ax+bf(x) = ax + b, the ϵ\epsilon-δ\delta proof follows a similar pattern, where δ\delta is typically ϵa\frac{\epsilon}{|a|}.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Epsilon-Delta Definition
Linear Functions

Formulas

lim x → a f(x) = L: For every ε > 0, there exists δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε
|(3x - 17) - 4| = 3|x - 7|

Theorems

Epsilon-Delta Definition of Limits

Suitable Grade Level

Grades 11-12