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?

As $\dfrac1{x^{1/n}}=\left(\dfrac1x\right)^{1/n}$, it suffices to establish the property for $x>1$.
It is enough to say
$$x^{1/n}-1=(x^{1/n}-1)\frac{x-1}{(x^{1/n})^n-1}=\frac{x-1}{(x^{1/n})^{n-1}+(x^{1/n})^{n-2}+\cdots1}<\frac{x-1}n.$$
