How can I prove by definition (with $\epsilon$ and $N$) that, for $c>0$,
$$\lim_{n \to \infty}\ \sqrt[n]{c} = 1 ?$$
Thanks.
How can I prove by definition (with $\epsilon$ and $N$) that, for $c>0$,
$$\lim_{n \to \infty}\ \sqrt[n]{c} = 1 ?$$
Thanks.
Assume first $c>1$ and write $c^{1/n}=1+h_n$ with $h_n>0$. Taking $n$-th powers and using the binomial theorem get $ c=(1+h_n)^n=1+nh_n+\dots>1+nh_n\implies 0
If $0
$\lim_{n\to\infty}\exp\left(\frac{\ln(c)}{n}\right)=\exp\left(\ln(c)\lim_{n\to\infty}\frac1{n}\right)=\dots$