Disprove the limit $\lim_{x\to 0}\frac{1}{x}=5$ with epsilon-delta

Question

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?

Answer 1

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 $\delta$.

Answer 2

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

Answer 3

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.