Prove that the limit does not go to 6 I want to prove that
$$ \lim_{x \to 2}  \ x + 3 \ne 6 $$
What I thought about doing was first assuming the limit actually equaled $6$. Then taking an $x$ below and above $3$ and then finding a contradiction form the two statements
1) choosing $x = 1.5$ we get $|0.5|< \delta \implies |0.5| < \epsilon$
2) choosing $x = 2.5$ we get $|0.5|< \delta \implies |1.5| < \epsilon$
Then choose an epsilon equal $1$ and use the contrapostive of $2$ but I am not sure exactly how to phrase this.
 A: Here is the definition of limit:
$\lim\limits_{x\rightarrow a} f(x)=L$ if for every $\epsilon>0$, there is a $\delta>0$ such that
$$|f(x)-L|<\epsilon\quad\text{whenever }\quad 0<|x-a|<\delta.$$
What does it mean that $\lim\limits_{x\rightarrow a} f(x)\ne L$?
Well, it means (imprecisely) that there is an $\epsilon>0$ such that $f(x)$ is not close to $L$ no matter how close $x$ is to $a$.
More precisely it means that there is an $\epsilon>0$, such that no matter how small $\delta>0$ is, there is an $x$ with $0<|x-a|<\delta$ and yet $|f(x)-L|\ge\epsilon$.
So, in your case, you need to find a fixed value of $\epsilon$ such that for any $\delta>0$ there is an $x$ such that the following holds:
$$
\tag{1}|(x+3)-6|\ge\epsilon\quad\text{and}\quad  0<|x-2|<\delta.
$$
Here, you could choose $\epsilon=1/2$.  
Given any $\delta>0$, choose any $x$ such that
 $0<|x-2|<\min\{\delta,1/2\}$. 
Then $x$ would be in the interval $(1.5,2.5)\,$. Consequently,
 $x+3$ would be in the interval $(4.5,5.5)$ and thus at least $1/2$ units away from 6.  That is, $|(x+3)-6|\ge1/2$.   
Informally, if $x$ is very  close to 2, then $x+3$ would be far away from 6. And so there 
would be no $\delta$ that "works" in the definition of limit. The quantity $x+3$ is at least 1/2 unit away from 6 whenever $x$ is within 1/2 of 2.
A: Write down the formal definition of limit: $\lim_{x\to 2} x+3 = 6$ means:
$$(\forall \epsilon>0)(\exists \delta>0)(\forall x)\Bigl[0<|x-2|<\delta \Rightarrow |(x+3)-6|<\epsilon\bigr]$$
Since you want to prove that this is not true, negate the statement:
$$\neg(\forall \epsilon>0)(\exists \delta>0)(\forall x)\Bigl[0<|x-2|<\delta \Rightarrow |(x+3)-6|<\epsilon\bigr]$$
Push the negation through the quantifiers to get
$$(\exists \epsilon>0)(\forall \delta>0)(\exists x)\Bigl[0<|x-2|<\delta \not\Rightarrow |(x+3)-6|<\epsilon\bigr]$$
Negating $P\Rightarrow Q$ yields $P\land \neg Q$, so the property to prove is
$$(\exists \epsilon>0)(\forall \delta>0)(\exists x)\Bigl[0<|x-2|<\delta \land |(x+3)-6|\ge\epsilon\bigr]$$
In other words, we need to find some $\epsilon$ such that for all $\delta$ there is an $x$ closer to $2$ than $\delta$ such that $|x+3-6|=|x-3|$ is larger than $\epsilon$. This $x$ is allowed to depend on $\delta$, but we must find an $\epsilon$ that works for every $\delta$.
Thus, it is wrong when in your argument you start by setting $x=1.5$ and $x=2.5$ without speaking of $\epsilon$ and $\delta$ first. Neither of these $x$'s can possibly work for $\delta=0.001$, for example.
Hint: $\epsilon = \frac 12$ works. Can you see why?
Note that when we negate the definition, the "burden of proof" reverses. When we want to show what the limit is, the adversary chooses an $\epsilon$, and we must then find a $\delta$ that works for every $x$ that the adversary picks afterwards. But when we want to show what the limit is not, we get to pick $\epsilon$ and (later) $x$, whereas the adversary tries to find a $\delta$ that will foil us.
A: If the limit is $6$ and $\varepsilon=1/2$, then there exists $\delta>0$ such that whenever $2-\delta<x<2+\delta$ and $x$ is not exactly $2$, then $6-\varepsilon<x+3<6+\varepsilon$.  That would mean $2.5<x<3.5$ whenever $2-\delta<x<2+\delta$ and $x\ne2$.  No matter what positive number $\delta$ is, there will be some numbers in $(2-\delta,2+\delta)$ that are not in $(2.5,3.5)$.  There you have a contradiction.
A: The statement $\lim\limits_{x \to 2} x + 3 = 6$ is equivalent to saying that for any $\epsilon>0$ we have some $\delta>0$ such that $0<|x - 2|<\delta\implies |x+3-6|<\epsilon$. In order to prove $\lim\limits_{x \to 2} x + 3 \neq 6$, we want to find some $\epsilon>0$ such that this is not true. Let's try $\epsilon = 1/2$. What we want is to find points $x_1,x_2,\ldots$ which are arbitrarily close to $2$ (so that we have some point $x_n$ such that $0<|x_n-2|<\delta$ for any $\delta>0$) such that $|x_n+3-6|>1/2$ for all $n$. How about $x_n = 2 +\frac{1}{n+2}$? Well, $|x_n - 2| = \frac{1}{2+n}$, so the points get arbitrarily close to $2$, and $|x_n+3-6| = 1-\frac{1}{n+2}>1/2$ for all $n$, so $\lim\limits_{x \to 2} x + 3 \neq 6$.
A: Let $L$ be some number other than $5$. Let's define $d=|L-5|$, and because $L\neq 5$ we have $d>0$. 
The reverse triangle inequality says that for any $a$ and $b$, 
$$|a-b|\geq ||a|-|b||.$$
In particular,
$$\left|L-(x+3)\right|=\left|(L-5)-\left(x-2\right)\right|\geq |d-|x-2||$$
Suppose that 
$$\lim_{x\to 2}\;(x+3)=L,$$
i.e. for any $\epsilon>0$, there exists an $\delta>0$ such that: for all $x$ with $|x-2|<\delta$, 
$$\left|L-(x+3)\right|<\epsilon.$$
Then we have that for all $x$ with $|x-2|<\delta$, 
$$|d-|x-2||<\epsilon$$
since $|d-|x-2||\leq|L-(x+3)|$. 
But $x=2$ will always have $|x-2|=0<\delta$ for any $\delta>0$, so in particular we must have
$$|d-|2-2||=|d|=d\leq\epsilon$$
for all $\epsilon>0$. But this is false, e.g. take $\epsilon=\frac{d}{2}$.
Thus, the limit cannot be any number other than 5, so the limit certainly cannot be 6.
A: There are seven steps to be followed when proving that the limit of f(x) as x->a is not = L, with the addition to negation shown above by David Mitra.
1. Find the real limit M for f(x) 
2. Choose epsilon so that real limit M is not in (L-eps., L+ eps.)
3. Write the definition in a negated form
4. Remove absolute value for f(x)-L >/= eps. , and find x intervals using the choosen eps.
5. Observe the interval where a falls in for delta>0, that will mean that the intersection of that interval with (a-delta, a+delta) is not =  empty set.
6. So whatever delta is,let x' be any point within an intersection.
7. Then absolute value of f(x')-L >/=eps. as required. Hence lim f(x) as x->a is not =L
