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?
$\lim_{n\rightarrow\infty} x^{1/n}=x^{\lim_{n\rightarrow\infty} 1/n}= x^0 = 1$ for any $x&gt;0$ (by continuity of the exponential function).
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 $:

$$\lim_{x \to \infty} f(x) = L \iff \forall_{ \epsilon &gt; 0}\ \exists_{\delta &gt; 0}: x &gt; \frac{1}{\delta } \Rightarrow |f(x) - L| &lt; \epsilon $$

Let us fix some $\epsilon &gt; 0$. 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 &gt; b \Rightarrow |f(x) - 1| &lt; \epsilon $.

Pausing for a moment we realize that for $x &gt; 1, \sqrt[x]{x} &gt; 1$ thus if $f(x) $ is to fall under a distance shorter than $\epsilon $ of 1, we must have $f(x) &lt; 1 + \epsilon $.

Now we just rewrite our inequality:

$$\sqrt[x]{x} &lt; 1 + \epsilon \iff x &lt; (1 + \epsilon)^x $$

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 $2 &lt; (1 + 0.01)^2$.

What one sees is that the inequality $x &lt; (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} $