If $x_n$ is a sequence converging to $r$, show that $x_n^{\frac{1}{n}}$ converge to 1 If $x_n$ is a sequence of positives numbers converging to $r>0$, show that $x_n^{\frac{1}{n}}$ converge to 1. I have problems with the proof, for example i tried this:
By hypothesis $x_n\rightarrow r$ then for all $\epsilon>0$ exist $N_{\epsilon}: |x_n-r|<\epsilon$. Now i want to estimate
$$|x_n^{\frac{1}{n}}-1|=|x_n^{\frac{1}{n}}-r^n+r^n-1|\leq |x_n^{\frac{1}{n}}-r^n|+|r^n-1|$$
i would like to estimathe the last inequelity for $\frac{\epsilon}{2}+\frac{\epsilon}{2}$. Other idea is to use $(a+b)^n=\sum_{i=1}^n \frac{n!}{(n-i)!i!}a^nb^{n-i}$ with $a=x_n^{\frac{1}{n}}$ y $b=-1$. Can somebody give me a hint please? Thank you
remark:In this moment i am thinking in perhaps helps: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})$.
 A: It can be done in a simpler way. As $\lim_{n\to\infty} x_n = r > 0$, then there exists $N \in \mathbb{N}$ such that
$$
\forall \; n \geq N: |x_n - r| \leq \frac{1}{2}{r} \Leftrightarrow 
\frac{1}{2} r \leq x_n \leq \frac{3}{2}r,
$$
which is just the definition of limit for $\varepsilon = \frac{1}{2}r$.
Hence $y_n = \ln x_n$ is defined at least for $n \geq N$, and also
$$
\ln\left(\frac{1}{2} r\right) \leq y_n \leq \ln\left(\frac{2}{2} r\right), n \geq N.
$$
As $\ln x_n^{1/n} = \frac{1}{n} \ln{x_n} = \frac{1}{n}y_n$, from
$$
\frac{1}{n} \ln\left(\frac{1}{2} r\right) \leq y_n \leq \frac{1}{n} \ln\left(\frac{2}{2} r\right), n \geq N
$$
and the squeeze theorem $\lim_{n\to\infty} \frac{1}{n} y_n = 0$, and by continuity of $\exp$ function
$$
\lim_{n\to\infty} x_n^{1/n} = \lim_{n\to\infty} e^{\ln x_n^{1/n}} = \exp\left\{\lim_{n\to\infty} \ln x_n^{1/n}\right\} = e^0 = 1.
$$
EDIT: another way to prove is to use the monotonicity of $f(x) = x^{1/n}$ and the fact that $\lim_{n\to\infty} a^{1/n} = 1$ for any $a > 0$, since
$$
\frac{1}{2} r \leq x_n \leq \frac{3}{2}r \Rightarrow
\left(\frac{1}{2} r\right)^{1/n} \leq x_n^{1/n} \leq \left(\frac{3}{2} r\right)^{1/n}.
$$
A: This isn't a hint but rather an alternative approach.
If $x_n \rightarrow r$ means that $\forall \epsilon > 0$ we can find a $\delta$ s.t. $\forall n > \delta$
$$ |x_n - r| < \epsilon$$
Since $\epsilon > 0$ This is equivalent to the condition that:
$$ x_n -r <  \epsilon  \\ r - x_n < \epsilon $$
[Assuming you have access to this inequality, perhaps you don't?] We know that if $a,b,c$ are positive then
$$ a^{\frac{1}{c}} < b^{\frac{1}{c}} $$
So we rewrite our two inequalities above as:
$$ x_n < \epsilon + r  \\ x_n  > r - \epsilon $$
And then we take exponents (and we can do this since $r>0$ so there must be SOME sufficiently small $\epsilon$ s.t. $r - \epsilon > 0$ and therefore the LHS and RHS of all sides of our equations are positive) to conclude that $\forall r > \epsilon > 0$ that  $\exists \delta$  s.t. $\forall n > \delta$
$$ x_n^{\frac{1}{n}} < \left( \epsilon + r \right)^{\frac{1}{n}} $$
$$ (r-\epsilon)^{\frac{1}{n}}< x_n^{\frac{1}{n}} $$
We can combine these then to state:
$$ (r-\epsilon)^{\frac{1}{n}}< x_n^{\frac{1}{n}} < \left( \epsilon + r \right)^{\frac{1}{n}}  $$
So if we can show that the LEFT MOST equation converges to 1 and the right most equation also converges to 1 then our middle must converge to 1 by the squeeze theorem.
