Math Problem Statement

Assuming 9 < x < 25 , you can choose delta = 7epsilon . Then, if 0 < | x- ( ) | < delta when delta = 7epsilon , you have | x- ( ) | < epsilon| sqrt x +4| | sqrt x -( 4)|< epsilon |f(x) - L| < epsilon

Solution

It seems like you're describing a limit problem, possibly involving the epsilon-delta definition of a limit. The setup implies that the function in question involves square roots, and you want to show that the function's value gets arbitrarily close to a limit LL as xx approaches a certain value.

Let's break it down:

Epsilon-Delta Definition:

The epsilon-delta definition of a limit states that for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that if 0<xc<δ0 < |x - c| < \delta, then f(x)L<ϵ|f(x) - L| < \epsilon.

In this case:

  • You seem to be dealing with a function that includes a square root, possibly f(x)=xf(x) = \sqrt{x}.
  • You want to prove that as xx approaches a certain value, the function approaches a limit LL.

Analyzing Your Statement:

  1. Bounds on xx: You are given 9<x<259 < x < 25, which suggests you're interested in xx values between 9 and 25.

  2. Delta Condition: You choose δ=7ϵ\delta = 7\epsilon, which means the delta depends linearly on epsilon. This is often done to make the expression manageable.

  3. Inequality Setup:

    • You have 0<x(?)<δ0 < |x - ( ? )| < \delta, where the missing part likely indicates the point xx is approaching.
    • The expression x(?)<ϵx+4x4|x - ( ? )| < \epsilon \cdot | \sqrt{x} + 4| \cdot | \sqrt{x} - 4| suggests that you are working with the difference between two square roots, which simplifies into a factored form, possibly involving the difference of squares.

    Specifically: f(x)L=x4<ϵ|f(x) - L| = |\sqrt{x} - 4| < \epsilon

    Here, L=4L = 4 might be the limit you are approaching, and the goal is to show that x\sqrt{x} gets close to 4 as xx gets close to 16.

  4. Conclusion: This suggests you're working to prove that limx16x=4\lim_{x \to 16} \sqrt{x} = 4.

Next Steps:

  • To complete this, you would determine δ\delta based on ϵ\epsilon to satisfy the inequality x4<ϵ|\sqrt{x} - 4| < \epsilon whenever 0<x16<δ0 < |x - 16| < \delta.

Would you like further details or clarifications on any part of this explanation?

5 Related Questions:

  1. How do you formally define the epsilon-delta definition of a limit?
  2. What is the significance of choosing δ\delta in terms of ϵ\epsilon in the epsilon-delta proof?
  3. How do you handle limits that involve square roots or more complex expressions?
  4. What are some common techniques for solving epsilon-delta problems involving inequalities?
  5. Can you provide an example of applying the epsilon-delta definition for a different function, like f(x)=x2f(x) = x^2?

Tip:

When proving limits with the epsilon-delta definition, always try to express δ\delta in terms of ϵ\epsilon to create a direct relationship between how close xx is to the target value and how close f(x)f(x) is to the limit.

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Math Problem Analysis

Mathematical Concepts

Limits
Epsilon-Delta Definition
Square Roots

Formulas

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Theorems

Epsilon-Delta Definition of a Limit

Suitable Grade Level

Advanced Mathematics