Prove that $|x|<1$ implies that $\lim_{n \to \infty} x^n=0$

Assuming you mean to show $\lim\limits_{n\to\infty} x^n=0$ here is a hint:

Without loss of generality assume $x\neq 0$, then there exists $h>0$ with $\dfrac{1}{|x|}=1+h$. Applying Bernoulli's inequality yields: $$\left|\frac{1}{x^n}\right|=\left(\frac{1}{|x|}\right)^n\geq 1+nh = 1+n\left(\frac{1}{|x|}-1\right)>n\cdot \frac{1-|x|}{|x|}.$$ Thus we have $$0\leq |x^n|<\frac{|x|}{1-|x|}\cdot \frac 1n$$

Now you should easily find such $N$ and $\varepsilon$ to use in the formal definition of a limit to prove $\lim\limits_{n\to\infty} x^n=0$.

Edit: although using the formal definition is not necessary as this would be a nice opportunity to apply the sandwich theorem.

This is another approach. If $x = 0$ then clearly the limit is $0$. Let us first consider $x$ to be positive so that $0 < x < 1$. Then clearly as $n$ increases the sequence $x^{n}$ decreases and is bounded below (by $0$). Hence $L = \lim\limits_{n \to \infty}x^{n}$ exists. Further we have $$L = \lim_{n \to \infty}x^{n + 1} = \lim_{n \to \infty}x\cdot x^{n} = x\lim_{n \to \infty}x^{n} = xL$$ or $(1 - x)L = 0$. Since $(1 - x) \neq 0$ it follows that $L = 0$. Hence for $0 < x < 1$ we have $$\lim_{n \to \infty}x^{n} = 0$$ If $-1 < x < 0$ then we put $x = -y$ so that $0 < y < 1$ and $x^{n} = (-1)^{n}y^{n}$. Since $y^{n} \to 0$ it follows that $x^{n} \to 0$ as $n \to \infty$.