Proof of $\lim_{x\to\infty}1/x=0$ I'm trying to prove that the limit of $1/x$ approaching zero as $x$ approaches $\infty$. 

My professor is asking that we use this definition: For all $\varepsilon&gt;0$ there exists $m$ such that if $x&gt;m$, then $∣f(x)−L∣&lt;\varepsilon$.

I made it to the point of For all $\varepsilon&gt;0$, if $x$ is greater than $\varepsilon$, then the absolute value of $1/x$ is less than epsilon. But I'm not sure if I went the right direction.
You wish to find an $m$ depending on $\varepsilon$ that makes the condition hold.

If $x&gt;\frac1\varepsilon$ then
$$\left|\frac{1}{x}-0\right|=\frac1x &lt; \frac{1}{\frac1\varepsilon} = \varepsilon, $$
thus you can choose $m = \frac1\varepsilon$.
The answer is a rephrasing of the original question: Is there a mathematical derivation to prove $\lim_{x \to \infty}\tfrac{1}{x} = 0$ without only assuming, guessing, or knowing the result and then confirming the result, $0$, using the definition of the limit, $L$?