Understanding the proof that $\lim_{n\to\infty}x^{1/n}=1$ ($x>0$) in Tao's Analysis I was reading sequences in Terence Tao Analysis book and I came across the question:

Prove that $\forall x>0$, $\lim\limits_{n\to\infty}x^{1/n}=1$

In the hint it says that

... you may need to treat the cases $x\geq1$ and $x<1$ separately. You might wish to use Lemma 6.5.2 to prove the preliminary result that $\forall\epsilon>0$ and every real number $M>0$, $\exists$ an $n$ such that $M^{1/n}\leq1+\epsilon$.
Lemma 6.5.2 : Let $x$ be a real number. Then $\lim_{n\to\infty}x^n$ exists and is equal to $0$ when $|x|<1$, exists and is equal to $1$ when $x=1$, and diverges when $x=-1$ or when $|x|>1$.

How does the above Lemma 6.5.2 help in proving the preliminary result given in the hint?And how does this preliminary result in turn help to prove our final result?

If $0 < x < 1$, then put $y = \dfrac{1}{x}$, we can consider only the case $x > 1$. By AM-GM inequality: $1 \leq \sqrt[n]{x}=\sqrt[n]{1\cdot 1\cdot 1\cdots \sqrt{x}\cdot\sqrt{x}}\leq \dfrac{(n-2)+2\sqrt{x}}{n}=1+\dfrac{2(\sqrt{x}-1)}{n}$.Next apply squeeze theorem to conclude.
