Prove $\lim_{n\to\infty}\frac{1}{\sqrt[3]{n}}=0.$ using the definition

Question

Prove $\lim_{n\to\infty}\frac{1}{\sqrt[3]{n}}=0.$ using the defenition

Let $\epsilon>0$ be arbitrarily given. Then pick $N \in \mathbb{N}$ where $N>\frac{1}{\epsilon}$. If $n \geq N$ then we have that(little unsure about this line) $n\geq \sqrt[3]{n}\geq N> \frac{1}{\epsilon}$. This implies that $|\frac{1}{n}-0|=\frac{1}{n} \leq \frac{1}{\sqrt[3]{n}} \leq N < \epsilon$. Therefore by definition $\lim\frac{1}{\sqrt[3]{n}}=0$.

Answer 1

How do you conclude that $\sqrt[3]{n} \ge N$? The correct version is:

Fix $\varepsilon>0$ and define $N:=1/\varepsilon^3$. Then, for all $n\ge N$, it holds $$ \left|\frac{1}{\sqrt[3]{n}}-0\right|=\frac{1}{\sqrt[3]{n}} \le \frac{1}{\sqrt[3]{N}} \le \frac{1}{\sqrt[3]{1/\varepsilon^3}}=\varepsilon. $$

Answer 2

It's good that you're unsure, because the line

$$n \ge \sqrt[3] n \ge N $$

is indeed completely wrong. All you've assumed is that $n \ge N$, not that $\sqrt[3]n \ge N$.

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So the fact that you can't say $\sqrt[3] n > \frac 1 {\epsilon}$ suggests that you should use a very different choice of $N$. Something like $$N = \frac{1}{\epsilon^{3}}$$ might be a bit more appropriate.

Answer 3

We must prove that$$\forall\epsilon>0,\quad\exists N\in\Bbb N,\quad\forall n>N,\quad |\dfrac{1}{n^{1\over 3}}|<\epsilon$$we try to find such $N$ for every such $\epsilon$. Here we start:$$|\dfrac{1}{n^{1\over 3}}|<\epsilon\to\\\dfrac{1}0<{n^{1\over 3}}<\epsilon\to\\\dfrac{1}{n}<\epsilon^3\to\\n>\dfrac{1}{\epsilon^3}$$therefore if we take $N=\lfloor\dfrac{1}{\epsilon^3}\rfloor+1$ we have proven what we want.