Math Problem Statement

In formally proving that lim_{x → 6} (x^2 + x) = 42, let ε > 0 be arbitrary. Choose δ = min(ε/m, 1). Determine the smallest value of m that would satisfy the proof.

Solution

The problem in the image asks to formally prove that:

limx6(x2+x)=42\lim_{x \to 6} \left( x^2 + x \right) = 42

Given that ϵ>0\epsilon > 0 is arbitrary, and with a choice of δ=min(ϵm,1)\delta = \min\left( \frac{\epsilon}{m}, 1 \right), the task is to determine the smallest value of mm that would satisfy the proof.

Step-by-Step Approach:

  1. Limit Setup: To prove limx6(x2+x)=42\lim_{x \to 6} (x^2 + x) = 42, we need to show that for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that if x6<δ|x - 6| < \delta, then (x2+x)42<ϵ|(x^2 + x) - 42| < \epsilon.

  2. Simplify the Expression: We start by simplifying (x2+x)42|(x^2 + x) - 42|: (x2+x)42=x2+x42|(x^2 + x) - 42| = |x^2 + x - 42| Factor the expression: x2+x42=(x6)(x+7)x^2 + x - 42 = (x - 6)(x + 7) Therefore: x2+x42=(x6)(x+7)|x^2 + x - 42| = |(x - 6)(x + 7)|

  3. Establish a Bound for x+7|x + 7|: Since x6x \to 6, we know x6<1|x - 6| < 1. Thus, xx lies between 5 and 7: 5<x<75 < x < 7 Therefore, x+7x + 7 lies between 12 and 14, which gives us the bound: 12x+71412 \leq |x + 7| \leq 14

  4. Choosing δ\delta: Now we need to ensure that (x6)(x+7)<ϵ|(x - 6)(x + 7)| < \epsilon. Using the bound on x+7|x + 7|, we have: (x6)(x+7)14x6|(x - 6)(x + 7)| \leq 14 |x - 6| To satisfy (x6)(x+7)<ϵ|(x - 6)(x + 7)| < \epsilon, it is enough to have: 14x6<ϵorx6<ϵ1414 |x - 6| < \epsilon \quad \text{or} \quad |x - 6| < \frac{\epsilon}{14}

  5. Conclusion: Thus, we choose m=14m = 14. The smallest value of mm that satisfies the proof is 14.

Final Result:

The smallest value of mm that would satisfy the proof is m=14m = 14.


Would you like more details or have any questions?

Here are five related questions to deepen your understanding:

  1. How would the limit change if the function were x2+2xx^2 + 2x instead of x2+xx^2 + x?
  2. Can you explain why we need to bound x+7|x + 7| in the proof?
  3. How would the choice of δ\delta change if mm were larger than 14?
  4. What happens to the limit if x7x \to 7 instead of x6x \to 6?
  5. Why do we consider the minimum of ϵm\frac{\epsilon}{m} and 1 when choosing δ\delta?

Tip: In epsilon-delta proofs, always simplify and factor the function as much as possible to clearly see how δ\delta relates to ϵ\epsilon.

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Math Problem Analysis

Mathematical Concepts

Limits
Epsilon-Delta Proof
Algebra

Formulas

lim_{x → c} f(x) = L (Epsilon-Delta Definition of a Limit)
(x^2 + x) - 42 = (x - 6)(x + 7)

Theorems

Epsilon-Delta Definition of a Limit

Suitable Grade Level

College level (Calculus I or Introductory Analysis)