# 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?

• If lim (1/x) =5 then by linearity lim (1/(5x))=1 . Substituting 5x=y gives lim(1/y)=1 . But this is absurd because this is the same function with two different limits at 0! – Jacob Wakem Aug 8 '16 at 20:31
• My solution is a complete rigorous solution. Please apply the bounty to it. – Jacob Wakem Aug 9 '16 at 12:38

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 $\delta$.

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.

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.

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.

====

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 $\lim_{x\rightarrow 0} 1/x \ne 5$.

• You want to say $0<|x|<\delta.$ – zhw. Aug 4 '16 at 11:37

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.

• Simple and clear! – paf Aug 9 '16 at 20:27

Yes, you can! Just show there's no $\delta$ corresponding to a specific value of $\epsilon$.

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 $\frac{1}{5+\varepsilon}$.

• – Scientifica Aug 9 '16 at 6:58
• @Scientifica It has been edited. Please upvote. – Jacob Wakem Nov 7 '16 at 20:26
• I'm not the one who downvoted your answer. – Scientifica Nov 7 '16 at 21:22