# How to prove that a limit is incorrect using epsilon delta definition of a limit?

I am trying to fully understand the Epsilon-Delta Definition of a Limit. I have no problem using it to prove a limit that is correct, but I am having trouble using it to disprove an incorrect limit.

For instance, I don't know how to disprove

lim(x -> 2) 2x = 40 It would help my understanding if someone could show me this (dis)proof.

• Just apply the definition and note that it requires $\forall x\, 0<|x-2|<\delta$ thus it must be valid for x "near" to 2. – gimusi Mar 23 '18 at 10:35

HINT

Recall that by definition $\forall \epsilon$ $\,\exists \delta$ such that $\forall x\,$, $0<|x-2|<\delta$ $\implies|2x-40|<\epsilon$

then

• take $\epsilon =2$
• $|2x-40|<2 \iff 19<x<21$

• then check the definition

• Sorry I didn't quite get it from check the definition. Can you explain why there isn't a delta for 19<x<21? – Star Platinum ZA WARUDO Mar 23 '18 at 13:18
• @ArjunNigam Note that for the $\epsilon$ adopted we need $19<x<21$ but the definition refer to $\forall x\,,0<|x-2|<\delta$ now take $x-2=1$ for example taht is $x=3$ and $|2\cdot 3-40|=34$ – gimusi Mar 23 '18 at 13:24
• I think I got it. Did you mean delta for ϵ=2 only exists for values of x between 19 and 21 but we need it to exist for all values of x? – Star Platinum ZA WARUDO Mar 23 '18 at 13:27
• @ArjunNigam The key point is that we need $\forall x\,,0<|x-2|<\delta$ thus choosing |x-2| sufficiently small we go out the bound fixed by $\epsilon$. Let try also with Others values. – gimusi Mar 23 '18 at 13:34
• Isn't that what I said? – Star Platinum ZA WARUDO Mar 23 '18 at 13:56

Let $\varepsilon=1$. Then, for every $\delta>0$, there is some $x\in\mathbb R$ such that$$0<|x-2|<\delta\text{ and }|2x-40|\geqslant 1=\varepsilon.$$For instance, take $x\in\bigl(2,2+\min\{1,\delta\}\bigr)$. Then $2<x<3$ and therefore $4<2x<6$. So, $|2x-40|\geqslant1$.

• shouldn't it be |2x-40|<=1? – Star Platinum ZA WARUDO Mar 23 '18 at 12:43
• @ArjunNigam No, since I am proving that the limit is not $40$. – José Carlos Santos Mar 23 '18 at 12:52