How to disprove $\lim_{x \to 2} 2x+5 =8$ using $\epsilon$ ,$\delta$ definition How to disprove $$\lim_{x \to 2} 2x+5 =8$$ using $\epsilon$ ,$\delta$ definition.
So for every $\epsilon$ we should not able to find $\delta$ such that we have $$|2x+5-8| \lt \epsilon$$ whenever $|x-2|\lt \delta$
So e have $$|2x-3|\lt \epsilon$$ that is
$$|2x-4+1|\lt \epsilon$$  or
$$\epsilon \gt |2x-4-(-1)| \gt |2x-4|-1$$
so
$$\epsilon \gt 2|x-2|-1$$  so
$$|x-2| \lt \frac{\epsilon+1}{2}$$
So now how can i say my $\delta$ is not $\frac{\epsilon+1}{2}$
 A: To negate the definition properly, we must show that there exists an $\epsilon>0$ such that for all $\delta>0$, there exists an $x$ such that $0<|x-2|<\delta$ but $|f(x)-8|\ge \epsilon$. Pay attention to the order of the quantifiers here: the limit definition is of the form
$$
\forall\epsilon\ \exists\delta\ \forall x: 0<|x-2|<\delta\implies|f(x)-8|< \epsilon
$$
so its negation should be
$$
\exists\epsilon\ \forall\delta\ \exists x: 0<|x-2|<\delta,\ \text{but}\ |f(x)-8|\ge \epsilon
$$
Hint: Try showing that if $\epsilon = 1/2$ then no matter what $\delta>0$ you pick, you can get $0<|x-2|<\delta$ but you will still have $|f(x)-8|\ge 1/2$.
A: To prove that the statement is false, I just have to pick a particular $\epsilon$ value, let me pick $\epsilon = 1$.
Suppose on the contrary that there is an $\delta> 0$, such that $|x-2| < \delta \implies |2x-3|<1$.
This is equivalent to $2-\delta < x < 2+\delta \implies 1 < x < 2$ which is not true.
In particular, $2+\frac{\delta}{2} \in (2-\delta, 2+\delta)$ but $2+\frac{\delta}{2} > 2$.
Remark:
Notice that while $|x-2| < \frac{\epsilon+1}{2} \implies \epsilon > |2x-4|-1$ and from triangle inequality we have $|2x-4-(-1)|\geq|2x-4|-1$, we cannot conclude that $\epsilon >|2x-4-(-1)|$
