so i got this 0<|x-a|<delta and |x^3-a^3|<epsilon and that is equal to |x-a||x^2+ax+a^2|<epsilon so i have to find a value that |x^2+ax+a^2|<value and |x^2+ax+a^2|=|x^2-2xa+a^2+3xa|<=|(x-a)^2|+|3xa| and i dont know what to do after this

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- Dec 10th 2012, 08:23 PMyayovioLim x^3= a^3 when x ->a by epsilon-delta definition?
so i got this 0<|x-a|<delta and |x^3-a^3|<epsilon and that is equal to |x-a||x^2+ax+a^2|<epsilon so i have to find a value that |x^2+ax+a^2|<value and |x^2+ax+a^2|=|x^2-2xa+a^2+3xa|<=|(x-a)^2|+|3xa| and i dont know what to do after this

- Dec 11th 2012, 07:01 AMhollywoodRe: Lim x^3= a^3 when x ->a by epsilon-delta definition?
You need to specify that $\displaystyle \delta$ is less than some fixed value, say 1. Then $\displaystyle |x^2+ax+a^2|<|(a+1)^2+a(a+1)+a^2|$, and I'll call the right-hand side M. So if you specify $\displaystyle \delta=\min(1,\frac{\epsilon}{M})$, you should be able to write out the proof with that.

There's a really good explanation of epsilon-delta proofs here at Math Help Forum - it's one of the four "sticky" threads at the top of the Calculus forum main page.

- Hollywood