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.

  • 2
    $\begingroup$ 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 $\delta$. $\endgroup$ – André Nicolas Jun 27 '14 at 17:30
  • $\begingroup$ even ignoring that I'm not sure what's going on with $"|x-4| < \delta \implies |(9-x-4)| < \epsilon"$ $\endgroup$ – John Fernley Jun 27 '14 at 17:32
  • $\begingroup$ 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. $\endgroup$ – Total Anime Immersion Jun 27 '14 at 17:32
  • $\begingroup$ Since I have taken 4 as the limit, |f(x) - 4| < € that's what I did. $\endgroup$ – Total Anime Immersion Jun 27 '14 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 $1/2$.

  • $\begingroup$ I think maybe a reason why "you will never find a $\delta$" is precisely the question. $\endgroup$ – Squirtle Jun 27 '14 at 17:36
  • $\begingroup$ Absolutely, but the question i was presented with, the one you can see above didn't have any E mentioned. $\endgroup$ – Total Anime Immersion Jun 27 '14 at 17:37
  • $\begingroup$ @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. $\endgroup$ – davidlowryduda Jun 27 '14 at 17:38
  • $\begingroup$ 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? $\endgroup$ – Total Anime Immersion Jun 27 '14 at 17:43
  • $\begingroup$ @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) $\endgroup$ – davidlowryduda Jun 27 '14 at 17:47

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