Epsilon-delta Proof: Show f(x)=sqrt(x) is Cont. @ x=4

[GBR]

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 :)

 

[losiu99]

This will work for sufficiently small ε. We have
|\sqrt{x}-2|=\frac{|\sqrt{x}+2|}{|\sqrt{x}+2|}|\sqrt{x}-2|=|x-4|/|\sqrt{x}+2|&lt;\frac{\epsilon^2}{|\sqrt{x}+2|}.
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|.

 

[Mark44]

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|.