Prove continuity by first principle

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Homework Help Overview

The problem involves proving the continuity of the function f(x) = x^2 at the point x = 2 using the ε - ∂ definition of continuity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the definition of continuity but struggles with manipulating the expression |x - 2||x + 2|. Some participants suggest introducing a restriction on |x - 2| to simplify the analysis.

Discussion Status

Participants are actively engaging with the problem, providing insights on how to express |x + 2| in terms of constants. A specific approach involving a limit as x approaches 2 is being explored, and a potential value for δ has been proposed.

Contextual Notes

There is a mention of an arbitrary restriction on |x - 2|, which is being used to facilitate the proof. The discussion reflects an ongoing exploration of how to effectively apply the ε - ∂ definition in this context.

chocolatefrog
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Homework Statement



Prove that f(x) = x^2 is continuous at x = 2 using the ε - ∂ definition of continuity.

2. The attempt at a solution

Using the definition of continuity, I've reached thus far in the question:

|x - 2||x + 2| < ε whenever |x - 2| < ∂

3. Relevant equations

I have no clue how to move forward from here. I know while solving this type of questions, we try to solve the first inequality so that ∂ can be written in terms of ε, but I can't seem to figure out what to do with |x + 2|.
 
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Okay, |x-2||x+2|&lt; \epsilon and so |x-2|&lt; \epsilon/|x+ 2|. That is almost what you want- you just need to replace that |x+2| with a constant. To do that, take some reasonable restriction on |x- 2|.

We want to take the limit as x goes to 2 so we really are only interested, say, in |x- 2|< 1 ("1" was chosen pretty much arbitrarily). That is the same as saying that -1<x- 2< 1 and so 3< x+ 2< 5. Since 3 itself is larger than 0, that is 3< |x+2|< 5. 1/5< 1/|x+2|< 1/3 so \epsilon/5&lt; \epsilon/|x+2|&lt;\epsilon/3. Now, what must your \delta be to guarantee that "if |x- 2|&lt; \delta, then |x- 2|&lt; \epsilon/|x+2|&amp;quot;?
 
Then \delta = \epsilon/5.
 
Oh, and I totally forgot, thanks a lot for your help! :)
 

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