# How to prove that a limit is wrong using Epsilon-Delta definition?

Suppose we have a function $f(x) = 9-x$. Now we know that $\lim\limits_{x\to4} f(x) = 5.$

Using the $\epsilon - \delta$ definition:

$|x-1| < \delta \implies |f(x) - 5| < \epsilon.$

What if we take a wrong limit now, lets say 4 for the given function like $\lim\limits_{x\to4} f(x) = 4$. We know its wrong but the epsilon delta defition still works, and I don't really understand how?

Like so :

$|x-4| < \delta \implies |(9-x-4)| < \epsilon$

$= |5-x| < \epsilon \equiv |x-5| < \epsilon$

Now we know if $\delta$ is $A$ and $\epsilon$ is $B$ then If $x < A \implies x < B$ then $B \leq A$.

Using this:

$|x-4| < \delta$ and hence $|x-4|$ is also less than epsilon. $|x-5| < |x-4|$ is also less than $\epsilon$ for all $x > 0$.

So here we have completed the proof which says to every $|x-x_0| < \delta$ there is a $|f(x) - L| < \epsilon$. Where is my mistake? Because the answer turns out to be wrong.

P.S. Someone please edit this document as I'm on a mobile phone and unable to do it.

• You need to show that for every $\epsilon$ there is a $\delta$ such that $\dots$. If you pick $\epsilon=1/10$, you can show there is no suitable $\delta$. Jun 27, 2014 at 17:30
• even ignoring that I'm not sure what's going on with $"|x-4| < \delta \implies |(9-x-4)| < \epsilon"$ Jun 27, 2014 at 17:32
• I guess the definition says for every |x-x0| < § implies |f(x) -L| < €. I did that in my above proof, but I wanna know exactly where I went wrong. Would be great help. Thank you. Jun 27, 2014 at 17:32
• Since I have taken 4 as the limit, |f(x) - 4| < € that's what I did. Jun 27, 2014 at 17:34

Remember the definition of limit, $$\forall \epsilon\; \exists \delta : |x-a|<\delta \rightarrow |f(x) - L|<\epsilon$$ Now if $L$ is not the limit negate the definition, that is $$\exists \epsilon\; \forall \delta: \;\; \exists x \text{ s.t. } |x-a| <\delta\text{ and } |f(x) - L| \geq \epsilon$$ In your case, as some people already mentioned take $\epsilon =1/2$, now for ever $\delta$ we need to find an $x$ such that $|x-4|<\delta$ and $|9-x-4|\geq 1/2$. Finding such $x$ isn't that difficult, just take $0<r<\min\{\delta, 1/2\}$ and consider $x=4+r$. To check we substitute in the inequalities: $$|x-4|=|r|<\delta$$ And $$|9-x-4|=|1-r|>1/2= \epsilon$$

(sorry for the awful format, I'm on my phone)

You haven't given $\delta$ or $\epsilon$ in your proof. This is what you need to do.

For instance, if you take $\epsilon = 1/2$, then you will never find a $\delta$ such that $\lvert x - 4\rvert < \delta \implies \lvert 9 - x - 4 \rvert < 1/2$.

One way to show this last statement is to see that one of the numbers $4.1, 4.01, 4.001, ...$ will be in the $\delta$-ball around $4$ no matter how small $\delta$ is, and for these $9 - x - 4$ is at least $0.9$, which is more than $1/2$.

• I think maybe a reason why "you will never find a $\delta$" is precisely the question. Jun 27, 2014 at 17:36
• Absolutely, but the question i was presented with, the one you can see above didn't have any E mentioned. Jun 27, 2014 at 17:37
• @Total: It is to be true for every $\epsilon > 0$. So if you can find any $\epsilon$ for which it doesn't hold, then it is not true. Jun 27, 2014 at 17:38
• If you take € to be 1/2 as you mentioned, you can solve the inequality for x, and do the same for the delta counterpart. Point is are we sure that the intervals wont match? Jun 27, 2014 at 17:43
• @Total: I rephrase my challenge. If you give me any $\delta$, I will give you back an $x$ within that range that makes your $\delta$ fail. (Truly, if you like. You give me the $\delta$, I'll respond with an $x$ value) Jun 27, 2014 at 17:47