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Epsilon-delta proof

  • Thread starter GBR
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  • #1
GBR
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Homework Statement



Use an epsilon-delta proof to show that f(x) = sqrt(x) is continuous in x = 4.


Homework Equations





The Attempt at a Solution



|f(x) - f(4)| = |sqrt(x) - sqrt(4)| < ε
= |x-4| < ε²

And now what? I'm not even sure I'm on the right path. My biggest problem during these proofs is choosing an ε and relating it to |x-a| < δ to complete the proof. Please bear with me; I recently went back to school as a physics undergrad, and realize my math-fu is destroyed after six years as a journalist :)
 

Answers and Replies

  • #2
133
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This will work for sufficiently small ε. We have
[tex]|\sqrt{x}-2|=\frac{|\sqrt{x}+2|}{|\sqrt{x}+2|}|\sqrt{x}-2|=|x-4|/|\sqrt{x}+2|<\frac{\epsilon^2}{|\sqrt{x}+2|}[/tex].
If, for example, ε<1, this is less than ε.
My strategy is to start with what you want to get (epsilon inequality) and work it backwards to get bound for |x-x0|.
 
  • #3
33,648
5,318

Homework Statement



Use an epsilon-delta proof to show that f(x) = sqrt(x) is continuous in x = 4.


Homework Equations





The Attempt at a Solution



|f(x) - f(4)| = |sqrt(x) - sqrt(4)| < ε
= |x-4| < ε²

And now what? I'm not even sure I'm on the right path. My biggest problem during these proofs is choosing an ε and relating it to |x-a| < δ to complete the proof. Please bear with me; I recently went back to school as a physics undergrad, and realize my math-fu is destroyed after six years as a journalist :)
You can't go from this inequality |sqrt(x) - sqrt(4)| < ε to this one |x-4| < ε². The reason is that |sqrt(x) - sqrt(4)|² is not equal to |x - 4|.
 

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