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 $\varepsilon>0$ there exists $m$ such that if $x>m$, then $∣f(x)−L∣<\varepsilon$.
I made it to the point of For all $\varepsilon>0$, 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.
 A: You wish to find an $m$ depending on $\varepsilon$ that makes the condition hold.
If $x>\frac1\varepsilon$ then
$$\left|\frac{1}{x}-0\right|=\frac1x < \frac{1}{\frac1\varepsilon} = \varepsilon, $$
thus you can choose $m = \frac1\varepsilon$.
A: I'm not sure that you're approaching this correctly. You claim that if $x>\varepsilon$ then $|\frac{1}{x}|<\varepsilon$. But what if $\varepsilon=\frac{1}{4}$? Then if $x=\frac{1}{2}$, we have that $x>\varepsilon$, but $$|\frac{1}{x}|= \ | \frac{1}{1/2}|=|2|>\varepsilon,$$ which clearly shows that the claim is false!
I suggest trying to find some rule for choosing $m$ so that whenever $x>m$ we have $|\frac{1}{x}|<\varepsilon$. For example, if $\varepsilon = 1$, what is the smallest $m$ satisfying the above property? What if $\varepsilon = \frac{1}{10}$? What if $\varepsilon=\frac{1}{100}$?
A: 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, $L$?
A: Let $\varepsilon >0$ be given; we must find a number $M$ such that for all $x$ with
 $x>M$ which  will imply $|1/x-0|=|1/x|<\varepsilon$.
The implication will hold if $M= 1/\varepsilon$ or any larger positive number. This proves that the limit as $x$ tends to $\infty$ of $1/x$ is equal to $0$.
