Use the definition to prove that $\lim_{n \rightarrow \infty} b^{\frac1n} = 1$, where $b>0$. Let $b > 0$. Use the deﬁnition to prove $\lim_{n \rightarrow \infty} b^{\frac1n} = 1$. 
please help me with that.
 A: Let $b>0$. Consider the following cases:
Case 1. $b=1$. The result is trivial.
Case 2. $b>1$. In this case, we need to show the following claim:

Claim: For all $n\in\Bbb N$ and $b>1$,  we have $b^{\frac{1}{n}}-1\leq \frac{b-1}{n}.$

Proof. For each $n\in\Bbb N$, write $d_n=b^{\frac{1}{n}}-1$. Because $b>1$, it follows that $d_n>0$ for each $n\in\Bbb N$. Thus, by using the Bernoulli inequality, we get
$$b=(1+d_n)^n\geq 1+nd_n\quad \forall n\in\Bbb N.$$ Hence for each $n\in\Bbb N$, we get
$$d_n\leq\frac{b-1}{n}$$
and this proves our claim.
We shall now prove the problem under case 2. Let $\epsilon>0$. Since $b>1$, we have $b-1>0$ and so $\frac{\epsilon}{b-1}>0$. By using the Archimedean Property, $\exists N\in\Bbb N$ such that $$\frac{1}{N}<\frac{\epsilon}{b-1}.$$ Thus, if $n\geq N$ then
$$\begin{align}
|b^{\frac{1}{n}}-1|&=b^{\frac{1}{n}}-1\\
&\leq\frac{b-1}{n}\\
&\leq\frac{b-1}{N}<\epsilon.
\end{align}$$
Hence, if $b>1$ then $$\lim_{n\to\infty}b^{\frac{1}{n}}=1.$$
This proves case 2.
Case 3. $0<b<1$. Then $\frac{1}{b}>1$ and then applying  Case 2, we get
 $$\lim_{n\to\infty}\bigg(\frac{1}{b}\bigg)^{\frac{1}{n}}=1,$$
that is,
$$\lim_{n\to\infty}\frac{1}{b^{\frac{1}{n}}}=1,$$
that is, if $0<b<1$ then
 $$\lim_{n\to\infty}b^{\frac{1}{n}}=1.$$
A: [HINT]
Case 1: 
$0<b<1$
We must prove there is an $n$ such that: 
$$\left|b^{\frac{1}{n}}-1\right|<\epsilon$$
For any $\epsilon>0$. We rearrange as such: 
$$-\epsilon<b^{1/n}-1<\epsilon$$
From here, the answer can be found through simple manipulation. 
Case 2: 
$b \geq 1$
Proving the case for $b=1$ will be easy, and for all $b > 1$, you can substitute $\frac{1}{b}$ in for the previous $b$ and apply case $1$. 
