Prove $\lim_{n\to\infty} 5^{\frac{1}{n}}=1$

Question

Prove

$\lim_{n\to\infty} 5^{\frac{1}{n}}=1$

Observe that we have \begin{align*} |5^{\frac{1}{n}}-1|&< \epsilon\\ 5^{\frac{1}{n}}&<\epsilon +1 \\ (\frac{1}{n})\ln{5}&<\ln{(\epsilon +1)}\\ \ln{5}&<n[\ln{(\epsilon +1)}]\\ \frac{\ln{5}}{\ln{(\epsilon +1)}}&<n \\ \end{align*}

Therefore let $\epsilon >0$ be arbitrarily given. Then choose $N> \frac{\ln{5}}{\ln{(\epsilon +1)}}$; then when $n\geq N$ that implies that

$|5^{\frac{1}{n}}-1|< \epsilon$ therefore $\lim5^{1/n}=1$.

However showing the last line i'm having trouble with now that I have my $N$.

Answer 1

I think your proof works.

Below is another proof.

Let $1+h_n=5^{\frac{1}{n}}$, then $h_n>0$ and $(1+h_n)^n=5$.

So, $\displaystyle 5>1+\binom{n}{1}h_n$ for $n>1$.

$\displaystyle h_n<\frac{4}{n}$.

For $\epsilon>0$, take $N\in\mathbb{N}$ such that $\displaystyle N>\frac{4}{\epsilon}$. Then for $n>N$,

\begin{align*} \left|5^{\frac{1}{n}}-1\right|&=h_n\\ &<\frac{4}{n}\\ &<\frac{4}{N}\\ &<\epsilon \end{align*}

Answer 2

It seems ok, you have found a value for $N$ such that for each $\epsilon>0$ for $n>N\quad |5^{\frac{1}{n}}-1|< \epsilon$, thus the limit is proved.