How do I prove that $\lim_{x\to \infty}(\sqrt[4]{x})$ is not Cauchy by using the definition of the Cauchy sequence

Question

The Cauchy Sequence is that: $\forall \varepsilon >0 \, \exists n_0 \in \mathbb ℕ$ such that $\forall m > n \geq n_0, \, |x_m−x_n|≤\varepsilon.$ Let's negate it, we will get $∃\varepsilon> \, ∀ \in \mathbb ℕ$ such that $∃>≥_0 \, |_−_|≥\varepsilon.$ Need to show that $|\sqrt[4]{_} - \sqrt[4]{_}| ≥\varepsilon $

Answer 1

You mean $x_n = \sqrt[4]{n}$, then on $[m, 2m], m > 1, f(x) = \sqrt[4]{x}$ is differentiable thus using MVT yields: $\left|x_m - x_{2m}\right| = \left|\sqrt[4]{m} - \sqrt[4]{2m}\right| \ge |2m - m|\cdot \dfrac{1}{4\sqrt{m}} = \dfrac{\sqrt{m}}{4} > \dfrac{1}{4} = \epsilon$. Thus it’s not Cauchy ( sequence ).