I'm trying to use the limit definition of a sequence to prove that the limit as $ n \rightarrow \infty$ of $\frac{1}{10^n}$ is equal to $0$. It is evident to see that this limit approaches 0,this is my work thus far.. not sure how I am allowed to rearrange for $n$ in terms of $\epsilon$ to complete the proof appropriately. I feel like it's something extremely basic.
If $\frac{1}{10^n} \rightarrow 0$, then $$\forall \epsilon > 0 , \exists N \in \mathbb{Z}$$ such that $ \lvert X_n - L \rvert < \epsilon$ whenever $ n \geq N$. Thus we let $ \epsilon > 0$, then $$ \lvert X_n - L \rvert = \lvert \frac{1}{10^n} - 0 \rvert$$ $$ \lvert X_n - L \rvert = \lvert \frac{1}{10^n} \rvert$$\ $$ \rightarrow 1/10^n < \epsilon$$
You're right that we need $\frac1{10^n}<\epsilon$ to make it work. To see when this can happen, note that we can equivalently write this as $10^{-n}<\epsilon.$ If you happen to know that the logarithm (base $10$) is an increasing function, then this is equivalent to $$-n<\log\epsilon,$$ which is in turn equivalent to $$n>-\log\epsilon.$$ Now use the Archimedean Property.
(Please let me know if I used any unfamiliar ideas, here.)
Perhaps a more elementary way, if you're familiar with induction, is to show that $n<10^n$ for all positive integers $n,$ so that $$\frac1{10^n}<\frac1n$$ for all such $n.$ Then all you have to do is make sure that $n>\frac1\epsilon$ and you're home free.