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?
 A: Let's assume that you know the proof for lemma 6.5.2. 
Assume on the contrary that for any $\epsilon > 0$ we have $M^{1/n} > 1 + \epsilon$ so that $(1 + \epsilon)^{n} < M$ for all $n$. Now $x = 1 + \epsilon > 1$ and hence $x^{n}$ diverges to $\infty$ and this contradicts that $x^{n} = (1 + \epsilon)^{n} < M$. Therefore for any $M > 0$ and any $\epsilon > 0$ there is a positive integer $n$ such that $M^{1/n} < 1 + \epsilon$. This establishes the preliminary result in the hint.

Now we consider the limit of $x^{1/n}$ when $n \to \infty$. If $x = 1$ then the limit is obviously $1$. Now suppose that $x > 1$. Then clearly $x^{1/n} > 1$ for all $n$. Let $\epsilon > 0$ be given. By preliminary result in last paragraph we know that there is a positive integer $m$ such that $x^{1/m} < 1 + \epsilon$. Note that $x > 1$ and hence if $n > m$ then $x^{1/n} < x^{1/m} < 1 + \epsilon$. It follows that for all values of $n \geq m$ we have $$1 - \epsilon < 1 < x^{1/n} < 1 + \epsilon$$ It is now clear (from definition of limit) that $x^{1/n} \to 1$ as $n \to \infty$.
For $0 < x < 1$ let us put $y = 1/x$ so that $y > 1$. Hence $y^{1/n} \to 1$ and $x^{1/n} = 1/y^{1/n} \to 1/1 = 1$ as $n \to \infty$.
Thus we see that for $x > 0$ we have $x^{1/n} \to 1$ as $n \to \infty$.
A: A slightly different approach:
First we'll show that $$\lim\limits_{n\to\infty} \sqrt[n]{n}=1$$ using the squeeze theorem.
For $n\in\mathbb N$ we have $1\leq\sqrt[n]{n}$. For $n\geq 2$ we can use the binomial theorem to get: $$n=\left(\sqrt[n]{n}\right)^n =[1+\left(\sqrt[n]n-1\right)]^n=\sum\limits_{k=0}^n \binom nk \left(\sqrt[n]n-1\right)^k\geq \binom n2 \left(\sqrt[n]n-1\right)^2.$$ Thus we have $$ \left(\sqrt[n]n-1\right)^2\leq \frac n{\binom n2}=\frac 2{n-1}.$$ As both sides of the equation are strictly positive we can take the root which yields $$\sqrt[n]n-1\leq \sqrt {\frac 2{n-1}} \Leftrightarrow \sqrt[n]n\leq 1+\sqrt{\frac 2{n-1}}.$$ As $\lim\limits_{n\to\infty} 1+\sqrt{\frac{2}{n-1}}=1$ we can now use the squeeze theorem to get $\lim\limits_{n\to\infty} \sqrt[n]{n}=1$.
Now: for $x>1$ there exists $n\in\mathbb N$ with $n\geq x$. This yields $$1\leq \sqrt[n]{x}\leq\sqrt[n]{n}$$ and we again apply the squeeze theorem.
For $0<x<1$ we have $\frac{1}{x}>1$ and thus we get $\lim\limits_{n\to\infty} \sqrt[n]{\frac{1}{x}}=1$ which also yields $\lim\limits_{n\to\infty} \sqrt[n]{x}=1$.
A: 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.
A: Other simplistic approaches $$A=x^{1/n}$$ $$\log(A)=\frac 1n \log(x)$$ So, $\log(A)\to 0$ and $A\to 1$.
Similarly, $$A=x^{1/n}=e^{\frac  {\log(x)} n}\to e^0=1$$
A: As $\dfrac1{x^{1/n}}=\left(\dfrac1x\right)^{1/n}$, it suffices to establish the property for $x>1$.
It is enough to say
$$x^{1/n}-1=(x^{1/n}-1)\frac{x-1}{(x^{1/n})^n-1}=\frac{x-1}{(x^{1/n})^{n-1}+(x^{1/n})^{n-2}+\cdots1}<\frac{x-1}n.$$
