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Artusartos
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Homework Statement
For question 19.2 in this link:
http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw7sum06.pdf [Broken]
I came up with a different proof, but I'm not sure if it is correct...
Homework Equations
The Attempt at a Solution
Let [tex]|x-y|< \delta[/tex]
For [tex]|f(x)-f(y)| = |x^2 - x^y| = |x-y||x+y| < \delta|x+y|[/tex], we know that the largest that |x+y| can be is 6. So if we let [tex]\delta= \epsilon/6 [/tex]...
We will have
[tex] |f(x)-f(y)| = |x^2 - x^y| = |x-y||x+y| < \delta|x+y| < (\epsilon/6)(6) = \epsilon [/tex]
If this is true for the largest possibility, then it must be possible for all of them...
Do you think my answer is correct, or is there something that I'm missing?
Thanks in advance
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