Disprove the limit $\lim_{x\to 0}\frac{1}{x}=5$ with epsilon-delta I understand how to prove a limit such as $\lim_{x\to 3}x^2=9$. Now I was wondering, can one also use the epsilon-delta method to disprove a limit such as:
$$\lim_{x\to 0}\frac{1}{x}=5$$
If so, how?
Thanks!
edit: what would a formal proof look like?

The $\varepsilon-\delta$ definition says:
$\displaystyle\lim_{x\to c} f(x)=L$ means: For all $\varepsilon>0$ there is some $\delta>0$ such that for all $x$ with $0<|x-c|<\delta$ we have $|f(x)-L|<\varepsilon$. 
So lets negate this: There is some $\varepsilon>0$ such that for all $\delta>0$ there is some $x$ with $0<|x-c|<\delta$ and $|f(x)-L|\geq \varepsilon$.
Now apply this to $\lim_{x\to 0}\frac{1}{x} \neq 5$.
Let $\varepsilon=420$. Consider any $\delta>0$. Then let $x=\min\{\delta/2,\frac{1}{425}\}$. Then in particular,  $\frac{1}{x}\geq 425$. Also note that $0<|x-0|<\delta$ and $|\frac{1}{x}-5|\geq420$ $\square$.
Basically, no matter how close we restrict ourselves to $0$, we can always escape the $\varepsilon$ of room we give ourselves (I picked $420$ just to pick something, you could pick any number really). So no matter how close we are to $0$ the function still blows up there. 
