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I´m having some trouble finding a delta for f(x)=(x-2)² using the epsilon-delta definition for fixed epsilon and x_0. Here´s what I come up with:

[tex]|f(x)-f(x_0)|<\epsilon[/tex]

[tex]|(x-2)^2-(x_0-2)^2|=|x^2-4x+4-x_0+4x_0-4|=|x^2-x_0^2-4(x-x_0)|=|(x-x_0)(x+x_0)-4(x-x_0)|=|x-x_0||x+x_0-4|<\epsilon[/tex]

now I divide by the second term and define my delta with the fixed epsilon and x_0, but there also appears this x which by definition should not be there, should it?

[tex]|x-x_0|<\frac{\epsilon}{|x+x_0-4|}:=\delta[/tex]

so, how can I make x disappear and thereby get some delta for my epsilon?

thanks a lot in advance!

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# Epsilon-delta test for continuity

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