Delta Epsilon Proof for Quadratic Limit

Introduction

• This lecture focuses on the delta epsilon proof for a quadratic limit.
• Differences from linear functions, as quadratics can have two answers.
• Aim is to keep the explanation as simple as possible.

Delta Epsilon Statement

• The statement for a limit is as follows:
  • If ( \lim_{x \to c} f(x) = l ), then for all ( \epsilon > 0 ), there exists a ( \delta > 0 ) such that:
    • If ( 0 < |x - c| < \delta ), then ( |f(x) - l| < \epsilon ).

Understanding the Proof

• We will prove this for the function ( f(x) = x^2 + 5x + 6 ) with ( l = 12 ) at point ( c = 1 ).
• We need to show:
  • ( 0 < |x - 1| < \delta ) implies ( |f(x) - 12| < \epsilon ).

Simplifying the Expression

• Start with:
  • ( |f(x) - 12| = |(x^2 + 5x + 6) - 12| = |x^2 + 5x - 6|.
• This can be factored into:
  • ( |(x + 6)(x - 1)| < \epsilon. )

Choosing ( x ) Values

• Need to determine a fixed distance for ( x ) around the target point (1).
• Suggestion is to keep ( |x - 1| < 1 ), leading to:
  • ( 0 < |x - 1| < 1 ) implies ( 0 < x < 2 ) and ( -1 < x < 1 ).
• This leads to the conclusion that:
  • ( |x + 6| < 8 ) when ( x ext{ is near } 1. )

Finalizing the Estimate

• Therefore, we can say:
  • ( |(x + 6)(x - 1)| < 8 |x - 1| < \epsilon. )
• Guess that ( \delta = \frac{\epsilon}{8} ) will satisfy the limit condition.

Conclusion and Part Two of the Proof

• In the second part of the proof, we verify:
  • Starting from ( |f(x) - 12| < 8\delta ) where ( \delta = \frac{\epsilon}{8} ) leads to:
  • ( |f(x) - 12| < \epsilon. )
• Final result is confirmed:
  • The function's distance from the limit can be made less than any ( \epsilon ) by appropriately choosing ( \delta ).

Final Thoughts

• This approach involves making reasonable assumptions and adjustments based on the behavior of quadratic functions.
• Encouragement to keep learning and exploring mathematical proofs.