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?

Other simplistic approaches $$A=x^{1/n}$$ $$\log(A)=\frac 1n \log(x)$$ So, $\log(A)\to 0$ and $A\to 1$.
Similarly, $$A=x^{1/n}=e^{\frac  {\log(x)} n}\to e^0=1$$
