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?
 A: 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. 
A: Let's assume $\lim_{x\rightarrow 0} 1/x = 5$.
Let $\epsilon = 1$.  Then there exist a $\delta$ so the $|x-0| = |x| < \delta \implies |1/x - 5| < \epsilon = 1$.  Let $x = \min(\delta/2, 1/6)<\delta$.  So $|x| < \delta$ so $|1/x - 5| < 1$.  But $x \le 1/6$. So $1/x \ge 6$.  So $|1/x - 5| \ge |6-5| = 1$.  This is a contradiction.
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More general.  For any $\epsilon > 0$ and $\delta > 0$.  Let $0 < x < \min(\delta, \frac 1{5+\epsilon})$.  Then $|x - 0| < \delta$ and $|1/x - 5| > |1/\frac 1{5+\epsilon} - 5| = |5 + \epsilon - 5| = \epsilon$.
So it is not the case for any $\epsilon > 0$ that there is a $\delta$ so that $|x - 0| < \delta \implies |1/x - 5| < \epsilon$.
So $\lim_{x\rightarrow 0} 1/x \ne 5$.
A: Given $\epsilon> 0$, we want to find $\delta> 0$ such that if $|x- 0|= |x|< |\delta|$ then $|\frac{1}{x}- 5|< \epsilon$.  Of course, $|\frac{1}{x}- 5|= |\frac{1- 5x}{x}|$ so, if x is positive, $|\frac{1}{x}- 5|<\epsilon$ is the same as $\frac{1- 5x}{x}< \epsilon$ or $1- 5x< \epsilon x$, $1< (5+ \epsilon)x$, $x> \frac{1}{5+ \epsilon}$.  But since the right hand side of that is positive, $|\frac{1}{x}- 5|<\epsilon$ cannot be true for all $|x|< \delta$ for any $\delta$.
A: It is sufficient to prove for positive values. Given $\varepsilon$, numbers small than $\frac{1}{5+\varepsilon}$ have image outside of $(5-\varepsilon, 5 + \varepsilon)$. Thus this contradicts there being any $\delta$ such that being less than $\delta$ implies being inside $(5-\varepsilon,5+\varepsilon)$ because there are always elements less than $\delta$ that are also less than $\frac{1}{5+\varepsilon}$.  
A: If $\lim_{x \to 0} \frac{1}{x} = 5$ then since $\lim_{x \to 0} x =0$ we have $\lim_{x \to 0} x  \cdot \frac{1}{x} = 1 = 0$ which is a contradiction. 
A: Hint:
What is the negation of
$$\forall \varepsilon>0\;\exists \delta>0\;\forall x\;\biggl(\lvert x\rvert<\delta\implies\biggl\lvert\frac1x-5\biggr\rvert<\varepsilon\biggr)?$$ 
Second hint:
Roughly said, the negation of an implication is a counter-example.
A: Yes, you can! Just show there's no $\delta$ corresponding to a specific value of $\epsilon$.
