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

We are asked to prove the limit:

\[
\lim_{x \to 7} (3x - 17) = 4
\]

### Preliminary analysis

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

\[
| (3x - 17) - 4 | < \epsilon
\]

#### Step 1: Simplify the expression \(|(3x - 17) - 4|\)
We start by simplifying the expression inside the absolute value:

\[
| (3x - 17) - 4 | = | 3x - 21 |
\]

Factoring out a 3 from the absolute value:

\[
| 3x - 21 | = 3 |x - 7|
\]

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

Now, for the limit to hold, we need:

\[
3 |x - 7| < \epsilon
\]

Dividing both sides of the inequality by 3:

\[
|x - 7| < \frac{\epsilon}{3}
\]

This suggests that we should choose \(\delta = \frac{\epsilon}{3}\).

### Proof

To formally prove the limit, we proceed as follows:

Let \(\epsilon > 0\) be given. Choose \(\delta = \frac{\epsilon}{3}\). Now, suppose \(0 < |x - 7| < \delta\). Then:

\[
|x - 7| < \delta = \frac{\epsilon}{3}
\]

Multiplying both sides by 3:

\[
3 |x - 7| < 3 \times \frac{\epsilon}{3} = \epsilon
\]

Thus, we have:

\[
|3x - 21| = 3 |x - 7| < \epsilon
\]

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

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

\[
\lim_{x \to 7} (3x - 17) = 4
\]

### Do you need any further details or have any questions?

Here are some related questions to expand on this topic:
1. How does the \(\epsilon\)-\(\delta\) definition of limits apply to nonlinear functions?
2. Can you explain the geometrical interpretation of the \(\epsilon\)-\(\delta\) definition of limits?
3. How would the \(\epsilon\)-\(\delta\) 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 \(\epsilon\)-\(\delta\)?
5. How can you use \(\epsilon\)-\(\delta\) limits to prove the continuity of a function at a point?

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

Prove the following limit. lim x → 7 (3x − 17) = 4

Prove the limit lim x → 7 (3x − 17) = 4 using the epsilon-delta definition of limits. This detailed step-by-step solution covers key calculus concepts for high school students in Grades 11-12, including limits, the epsilon-delta definition, and linear functions.

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