limit as $x$ approaches infinity of $\frac{1}{x}$

What you are trying to prove is that "if $x$ is large enough, then $\frac{1}{x}$ gets close enough to $0$." Approaching infinity represents the concept of allowing $x$ to be arbitrarily large.

Here is how to state the proof, with some comments along the way. Let $\varepsilon>0$ be some arbitrary real number (this will represent "error" or "distance from 0"). Let $N=\frac{1}{\varepsilon}$ (this represents how close to infinity $x$ must be). Whenever $x>N$, then $\frac{1}{x}<\frac{1}{N}$, and since $\frac{1}{N}=\varepsilon$ we have $\left|\frac{1}{x}-0\right|<\varepsilon$ (that is, the distance between $\frac{1}{x}$ and $0$ is less than the error $\varepsilon$). Therefore, $\lim_{x\to\infty}\frac{1}{x}=0$.

Let $f:\Bbb R\setminus\{0\}\to\Bbb R$ be $f(x)=1/x$. You have conjectured that $$ \lim_{x\to\infty}f(x)=0 $$ Rigorously, this means that for every $\varepsilon>0$ there exists an $M$ such that $$ \left\lvert\frac{1}{x}\right\rvert<\varepsilon $$ whenever $x>M$. Can you prove this?