Does $f_n(x) = x^{1/n}$ converge? How would you prove that $f_n(x) = x^{1/n}$ converges pointwise to $1$ for $x \in (0, \infty)$? It's obvious if you take the limit as $n \to \infty$ but how would you prove it using epsilon?
 A: $\lim_{n\rightarrow\infty} x^{1/n}=x^{\lim_{n\rightarrow\infty} 1/n}= x^0 = 1$ for any $x>0$ (by continuity of the exponential function).
A: If $x^{1/n} \to 1$ pointwise then:
For all $\varepsilon > 0$, there exist an $N_x > 0$ such that $|x^{1/n}-1|<\varepsilon$ for all $n > N_x$.
Let's split this in to two cases: $x > 1$ and $0 < x < 1$.
If $x>1$ then $x^{1/n} > 1$ for all $n \in \mathbb N$, and hence $x^{1/n}-1>0$.
If $x^{1/n}-1>0$ then: $|x^{1/n}-1|<\varepsilon \iff x^{1/n}-1<\varepsilon$.
Let's try to find an $N$ for which $x^{1/n}-1<\varepsilon$, for all $n > N$.
If $x^{1/n}-1 < \varepsilon$ then $x^{1/n} < \varepsilon+1$. We can log both sides, and since log is an increasing function, it preserves the direction of the inequality.
\begin{eqnarray*}
\log(x^{1/n}) &<& \log(\varepsilon+1) \\ \\
\frac{1}{n}\log x &<& \log(\varepsilon+1) \\ \\
\frac{1}{n} &<& \frac{\log(\varepsilon+1)}{\log x} \\ \\
n &>& \frac{\log x}{\log(\varepsilon+1)}
\end{eqnarray*}
Notice dividing by $\log x$ didn't change the direction of the inequality because $x>1$ and $\log x > 0$.
Can you do the same for $0<x<1$?
A: The user @mathematician already gave a proof using limit's properties, let us now prove it by the epsilon-delta definition of limit.
I will start by writing it for the particular case when $x \to \infty $:
$$\lim_{x \to \infty} f(x) = L \iff \forall_{ \epsilon > 0}\ \exists_{\delta > 0}: x > \frac{1}{\delta } \Rightarrow |f(x) - L| < \epsilon $$
Let us fix some $\epsilon > 0$. We want to show that $f(x) \to 1$ as $x \to \infty $.
The condition for the existance of $\delta $ is equivalent to finding a lower bound $b $ such that $x > b \Rightarrow |f(x) - 1| < \epsilon $.
Pausing for a moment we realize that for $x > 1, \sqrt[x]{x} > 1$ thus if $f(x) $ is to fall under a distance shorter than $\epsilon $ of 1, we must have $f(x) < 1 + \epsilon $.
Now we just rewrite our inequality:
$$\sqrt[x]{x} < 1 + \epsilon \iff x < (1 + \epsilon)^x $$
This is trivially satisfied by $x = 1$, for any $\epsilon $, but surely $b \not = 1$ for any $\epsilon $. For example, taking  $\epsilon = 0.01$, surely we don't have $2 < (1 + 0.01)^2$.
What one sees is that the inequality $x < (1 + \epsilon)^x $ is satisfiable for very small $x $, and then for very big values of $x $. This means your lower bound $b $ is the value of $x $ for which any value above satisfies given inequality.
As a small example, we set $\epsilon = 0.001$ and ask mathematica to find our answer:

meaning your lower bound is $9123.13$ or that your delta is $\delta = \frac{1}{9123.13} $
