Showing that a limit is false, did I prove it? I want to show that $\lim_{x \to 2} x = 5$ is false. 
Let $\epsilon > 0$. If this limit equivalence were true, we would need a $\delta$ such that $|x - 5| < \epsilon$ whenever $0 < |x-2| < \delta$.
$|x - 5| < \epsilon$ can be rewritten to $3-\epsilon < x - 2 < 3 + \epsilon$ or $|x-2| < \min(3 + \epsilon, \epsilon - 3) = \epsilon - 3$
However, we cannot set $\delta = \epsilon - 3$, since for $\epsilon \leq 3$, we have $\delta \leq 0$, which does not satisfy our original requirement that $0 < |x-2| < \delta$.
 A: By contradiction suppose that
$$\lim_{x \to 2} x = 5$$
then by definition of limit we have that $\forall \epsilon>0$ $\exists\delta>0$ such that $\forall x$
$$0<|x-2|<\delta \implies |x-5|<\epsilon$$
but for $ \epsilon=1$  
$$|x-5|<1\iff 4<x<6$$
and for $\delta<1$
$$|x-2|<1\iff 1<x<3$$
then
$$\lim_{x \to 2} x \neq 5$$
Note that from the logical point of view since the definition is
$$\left(\lim_{x\rightarrow a} f(x) = L\right) \iff \Big(\forall \varepsilon >0\, \exists \delta > 0: \big(0<\vert x-a\vert <\delta \implies \vert f(x)-L\vert <\varepsilon\big)\Big)$$
the negation is
$$\left(\lim_{x\rightarrow a} f(x) \neq L\right) \iff \Big(\exists \varepsilon >0\, \forall \delta > 0: \big(0<\vert x-a\vert <\delta \not\Rightarrow \vert f(x)-L\vert <\varepsilon\big)\Big).$$
A: $0 < |x-2| < \delta$ can be rearranged to $x \neq 2$ and $-3-\delta < x-5 < \delta - 3$. 
$|x-5| < \epsilon$ can be rearranged to $-\epsilon < x-5 < \epsilon$
If the former condition implied the latter, we'd have $-\epsilon \leq -3-\delta$ and $\delta - 3 \leq \epsilon$.
This simplifies to $\epsilon \geq 3 + \delta$.
Since we are after the negation of this, we have:
$\epsilon < 3 + \delta$
Since $\delta > 0$, this suggests any $\epsilon \leq 3$ should prove the limit false regardless of $\delta$, i.e. you won't be able to find a $\delta$ that works. Unless I made a mistake with this somewhere.
A: Let $\epsilon_0:=1$. Given any $\delta>0$ put
$x:=2-{\delta\over2}$. Then $|x-2|<\delta$, but  $|x-5|\geq1$. It follows that no $\delta>0$ whatsoever suffices for the given $\epsilon_0$.
