There is another post showing a different solution to this problem, but I wanted to try using the delta-epsilon definition to show that as $n$ tends to infinity, $x^n$ converges to $0$ for $|x| < 1$. Please correct or suggest correction -
Let $\epsilon >0$ be given, and let $\delta = \min \{ \epsilon^{1/n}, 1 \}$
Then
$|x^n - 0| = |x^n| = |x|^n < \delta^n <= (\epsilon^{1/n})^n = \epsilon$
My apologies on the formatting. To be clear above, I'm referring the nth-root of epsilon whenever appropriate.
As Alann Rosas points out in the comments, there's a big problem with your proof, and it is not to do with the $\delta < 1$. The issue is that you are actually showing $\lim_{x \to 0} x^n = 0$, not $\lim_{n \to \infty} x^n = 0$. To show the latter, you need to use the $\varepsilon$-$N$ definition of a limit. That is, for any $\varepsilon > 0$, there would have to exist some $N$ such that $$n > N \implies |x^n| < \varepsilon.$$ There's a kind of "obvious" way of proceeding here: take the natural logarithm of both sides of $|x^n| < \varepsilon$ to get the equivalent inequality: $$n \ln |x| < \ln \varepsilon \iff n > \frac{\ln \varepsilon}{\ln |x|}.$$ (Note: $\ln|x| < 0$ as $|x| < 1$.)
The "obvious" way has a very subtle pitfall here, to do with the definition of the logarithm. It depends how the logarithm is specifically defined (there are several equivalent definitions), but often the fact that $x^n \to 0$ as $n \to \infty$ is used somewhere in the construction of the logarithm. In many treatments of real analysis, this makes the proof lamentably circular!
Fortunately, there is another proof that avoids this pitfall, and it's not too bad. If we consider the sequence $a_n = |x|^n$, then we can represent it as a recurrence relation: $$a_{n+1} = |x|a_n, \quad a_0 = 1.$$ Because $0 < |x| < 1$, we have $0 < a_{n+1} < a_n$, just by the usual ordered field axioms. This makes $a_n$ both bounded below by $0$ and monotone decreasing, so the monotone convergence theorem proves that a limit $L$ does exist. If we take the limit of both sides of the recurrence relation, we get $$L = |x|L \implies L(1 - |x|) = 0 \implies L = 0,$$ as $1 - |x| \neq 0$. This proves the limit is $0$, as required.