How to show that some number is not a limit of a function by the negation of the definition of limits The negation of the definition of limit of functions is $\exists\epsilon>0\forall\delta>0\exists x\neq a(|x-a|<\delta \text{ and } |f(x)-f(a)|\geq\epsilon)$. How do we use this to show for example that $lim_{x\rightarrow 3}x\neq 6$ or $lim_{x\rightarrow 3}x^2\neq 18$?
From the first limit we have that the following must apply $\exists\epsilon>0\forall\delta>0\exists\neq 3(0<|x-3|<\delta \text{ and } |x-6|\geq\epsilon)$. I get stuck at $\forall\delta>0$ since for me it implies that $x\in\mathbb{R}$ but $x\neq 3$ and with this there can not be any $\epsilon>0$ such that $|x-6|\geq \epsilon$.
Edit: So the negation of the definition should be $\exists\epsilon>0\forall\delta>0\exists x\neq a(0<|x-a|<\delta \text{ and } |f(x)-f(a)|\geq\epsilon)$ I think. I'm going to try to work with this but I appreciate answers.
Edit 2: This doesn't seem trivial at all. If we want to show that the negation of the definition applies to the first limit then we need to show that that there exists a $\epsilon>0$ such that for all $\delta>0$ there is a $x\neq 3$ such that the conditions are met. But the problem for me is $\delta>0$, with that I can't pick a $x$ since I can always make $\delta$ smaller such that the $x$ is not valid anymore.
 A: Why is $\lim_{x\to 3} x \ne 6$?
As $x$ gets very very close to $3$ we do not get very close to $6$.  We need to find a range were we can always pick an $x$ close to $3$ where $6-\epsilon < x < 6 + \epsilon$ never happens.
That is we must pick an $\epsilon$ were there is some cluster around $3$ that completely avoids the range from $6-\epsilon$ to $6+\epsilon$.
Well if $\epsilon = 2$ then we can make $6-2=4 < x < 6+2 = 8$ never happen by restricting our choice of $x$ to not ever but more than $4$.
So if $\delta = 1$ and we pick $|x-3|< 1$ then $2< x < 4$ and so $|x-6| \ge 2$.  So the $\lim_{x\to 3}x \ne 6$.
But wait a minute!  I hear you ask.  I thought it was true for all $\delta$; not just $\delta = 1$.
Well it needs to be true that for all $\delta$ you can pick one $x$ that fails.  If $\delta > 1$ then just pick an $x$ where $|x-3| < 1< \delta$.  If $\delta < 1$ then you can pick any $x$.  But for any $\delta$, even $\delta = 2$ billion. You can pick an $x$ in that range that is much closer to $3$ so it fails.
How would you write this up as a proof?
Pf:  Let $\epsilon = 2$.  Then for any $\delta > 0$, if we choose an $x$ so that $|x-3| < \min(1, \delta)\le \delta$ we will have $|x-3| < 1$ and therefore $2 < x < 4$ and therefore $|6-x| > 2 =\epsilon$ ($2<x<4$ so $-4<x-6<-2$ so $|x-6| > 2$)
and the definition fails. There is no $\delta > 0$ so that $|x-2| < \delta$ must imply $|6-x| < 2$.
.... and to prove
$\lim_{x\to 3}x^2 \ne 18$.
Proof:  Let $\epsilon =1$.  Then for any $\delta> 1$ pick an $x$ so that $|x-3| < \min(1,\delta)$.  Then $|x-3|< 1$ and $2< x<4$ and so $4 < x^2 < 16$.  Therefore $|x^2 -18| \ge 2 > 1=\epsilon$.
The definition of limit fails.
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Perhaps a less trivial example is to prove $\lim_{x\to 0}\cos \frac 1x$ does not exist.
Pf:  First note that for all $\delta > 0$ then $\frac 1{\delta}$, although possibly quite large, is finite and there will always exist some $2n\pi;n\in \mathbb N$ and $(2n+1) \pi;n\in \mathbb N$ where $\frac 1\delta < 2n\pi < (2n + 1)\pi$.  ANd that means: for any $\delta$ there will always exists $x_0 = \frac 1{2n\pi}$ and $x_1 =\frac 1{(2n+1)\pi}$ where $0 < x_1 < x_0$ and $|x_i - 0|=x_i < \delta$.
Now let $\epsilon = 1$.
Let $M\le 0$ be any non-positive real number.  If we select $x_0$ we have $|x_0 - 0| < \delta$ but $|\cos \frac 1{x_0} - M|=|\cos 2n\pi-M| = |1-M| = |1+|M|| \ge 1=\epsilon$.
Therefore, by definition, $\lim_{x\to 0}\cos\frac 1{x} \ne M$ for any $M \le 0$.
And let $K \ge 0$ be any non-negative real number.  If we select $x_1$ we have $|x_1-0| < \delta$ but $|\cos \frac 1{x_1} - K| = |\cos (2n+1)\pi -K| = |-1-K| = |1+K| \ge 1 = \epsilon$.
Therefore, by definition, $\lim_{x\to 0}\cos\frac 1x \ne K$ for any $K \ge 0$.
So $\lim_{x\to 0}\cos \frac 1x$ existing is impossible (assuming we are talking about real numbers, which are all either non-negative or non-positive [or both in the case of $0$])
A: So the first limit is $lim_{x\rightarrow 3} x=6 $ and we need to show that $\exists\epsilon>0\forall\delta>0\exists x\neq 3(0<|x-3|<\delta \text{ and } |x-6|\geq\epsilon)$.
Let $\epsilon=2$. For $0<\delta\leq1$ we have the interval $(3,4)$ or $(2,3)$ to choose the $x$ from. If $x\in (3,4)$ then $2 < |x-6|< 3$ and we can see that $\epsilon= 2$ holds. For $\delta > 1$ we can choose $x=4$ which we meets the condition of $\epsilon$. Thus we have proven that $|x-6|\geq 2$ for all $\delta>0$.
This looks correct but I'm not sure.
