Prove the statement using e and s

Question

lim          (1-4x)=13

x→-3

 

Prove the statement using ε and s

please show steps and similar example

Answer 1

I'll prove this limit statement using the formal epsilon-delta (ε-δ) definition of a limit.

Proof that lim(x→-3) (1-4x) = 13

Statement to prove: For every ε > 0, there exists a δ > 0 such that if 0 < |x - (-3)| < δ, then |(1-4x) - 13| < ε.

Proof:

Step 1: Start with what we need to make small: |(1-4x) - 13| < ε

Step 2: Simplify the left side: |(1-4x) - 13| = |1 - 4x - 13| = |-4x - 12| = |-4(x + 3)| = |-4| · |x + 3| = 4|x + 3|

Step 3: So we need: 4|x + 3| < ε

Step 4: Divide both sides by 4: |x + 3| < ε/4

Step 5: Notice that |x + 3| = |x - (-3)|, which is exactly the distance from x to -3.

Step 6: Choose δ = ε/4

Step 7: Verify the proof works: If 0 < |x - (-3)| < δ, then:

1. |x + 3| < ε/4
2. 4|x + 3| < 4(ε/4)
3. 4|x + 3| < ε
4. |(1-4x) - 13| < ε ✓

Conclusion: For any ε > 0, we can choose δ = ε/4, and the limit condition is satisfied. Therefore, lim(x→-3) (1-4x) = 13. ∎

Similar Example: Prove lim(x→2) (3x + 1) = 7

Step 1: Start with what we need: |(3x + 1) - 7| < ε

Step 2: Simplify: |(3x + 1) - 7| = |3x + 1 - 7| = |3x - 6| = |3(x - 2)| = 3|x - 2|

Step 3: We need: 3|x - 2| < ε

Step 4: Solve for |x - 2|: |x - 2| < ε/3

Step 5: Choose δ = ε/3

Step 6: Verify: If 0 < |x - 2| < δ, then:

1. |x - 2| < ε/3
2. 3|x - 2| < ε
3. |(3x + 1) - 7| < ε ✓

Answer: δ = ε/3

Key Pattern for Linear Functions

For limits of the form lim(x→a) (mx + b) = L:

1. Simplify |(mx + b) - L| to get |m||x - a|
2. Set |m||x - a| < ε
3. Choose δ = ε/|m|