$\delta$ = min {1, $\epsilon$} works for proving $\lim_{x->0}$ $x^3$ = 0?

Question

I'm quite sure that $\delta$=$\epsilon^{1/3}$ works well,

but I'm not sure $\delta$ = min {1, $\epsilon$} works as well.

If $\delta$ = min {1, $\epsilon$}, then in the case of $\epsilon$ < 1

|$x^3$| = $x^2$|x| < |x| < $\epsilon$.

And similarly for the case of $\epsilon$ ≥ 1 where $\delta$ = 1.

Does this argument work?

Note that if δ works then every δ'<=δ works, and that min{1,ϵ}<=ϵ^{1/3} for every ϵ. — Did, Jan 12 '15 at 07:24
@Rustyn ?? How do you understand "f(ϵ)<g(ϵ) for every ϵ"? — Did, Jan 12 '15 at 07:41

score 1 · Accepted Answer · answered Jan 12 '15 at 07:34

1

Your work is correct. There are many possibilities for $\delta$, in fact, if $\delta$ is a number that works, then any number smaller than $\delta$ will also work. This follows directly from the definition of the limit, i.e. $|x < c| < \gamma$ holds whenever, $\gamma \le \delta$.

answered Jan 12 '15 at 07:34

MathMajor

6,478

No problem, happy to help! – MathMajor Jan 12 '15 at 07:39

$\delta$ = min {1, $\epsilon$} works for proving $\lim_{x->0}$ $x^3$ = 0?

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