# Delta-Epsilon Proof for Cube Root Function [closed]

I need to find following limit and prove it using the delta-epsilon definition of the limit: $$\lim \limits_{x \to 0}x^{1/3}$$ I am familiar with using the delta-epsilon proof for linear and quadratic functions, but not for radicals. All help is appreciated.

## closed as off-topic by Did, Namaste, C. Falcon, Adam Hughes, user21820Feb 22 '17 at 10:41

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Let $\epsilon > 0$. Then, note that there exists $X >0$ such that $X < \epsilon^3$ by the Archimedes principle. But $X < \epsilon^3 \implies |X^{\frac{1}{3}}| < \epsilon$, and for any $y$ such that $|y|<X$, $|y^{\frac{1}{3}}| < |X^{\frac{1}{3}}| < \epsilon$, because the cube root function is the inverse of the cube function which is an increasing function, hence it is also always increasing. Thus, whenever $|x|<X$, $|x^{\frac{1}{3}}| < \epsilon$. Hence it follows that $\displaystyle\lim_{x \to 0} x^\frac{1}{3} = 0$.

• Yes, the crucial thing one needs to prove here is that for any $n\in\Bbb N$, $0<x<y \implies x^{1/n}<y^{1/n}$. – Ted Shifrin Aug 25 '16 at 23:51
• @TedShifrin I've added a small explanation. I think the questioner should be able to figure out my logic now. – астон вілла олоф мэллбэрг Aug 25 '16 at 23:56