Question about limits using delta - epsilon definition So I was trying to understand better limits using the $\epsilon$-$\delta$ definition and I decided to disprove a limit which is obviously false.
Say, $\lim_{x\to 1} (x + 2) = 10$. Ok, this is false because the limit is  $3$, but using the definition:
$|(x+2) - 10|< \epsilon$ (this is my epsilon)
$|(x-1)| < \delta$ (this is my delta)
$|x-1 -7| < \epsilon$
$|x-1|< \epsilon+7$
so it looks like if my $\delta$ is equal to $\epsilon +7$, I should get an $\epsilon$ bigger than $\delta$, since the limit is false. But if $\epsilon$ is say $20$, $\delta$ can be at most $27$, so let us pick $x$ to be $26$
$|(26-2) - 10| < 20$
$|24 - 10| < 20$
$|14| < 20$
So it works!! so my question is, where is my mistake?
 A: The definition of limit is given in the form of a theoretical test (meaning  that in general it requires more of thought than calculations to apply it) to check whether or not a given number $L$ is the limit of a function $f$ as $x \to a$.
You want to use the definition of limit to test whether $L = 10$ is the limit of $f(x) = x + 2$ as $x \to a = 1$. For simple problems the definition of limit is well suited to test these things. Intuitively you know that $L = 10$ is not the desired limit and hence you expect the test given in definition of limit to fail.
In order to understand when the test fails you need to clearly understand when the test passes and then simply negate those conditions. But it is tricky to negate complex logical statements and requires some patience. Let's start with test of a limit.

A) $L = \lim_{x \to a}f(x)$ if "for every $\epsilon > 0$ there exists a $\delta > 0$" such that "$|f(x) - L| < \epsilon$ for all $x$ satisfying $0 < |x - a| < \delta$".

The statement above has two main parts and first part requires the existence of one $\delta > 0$ for every $\epsilon > 0$". So if we negate this we achieve the following:

B) "For some particular $\epsilon > 0$ there is no desired $\delta > 0$".

What is the desired $\delta$? It is one which allows the second part of the test. Thus a desired $\delta$ is one which makes the following statement

C) $|f(x) - L| < \epsilon$ for all $x$ satisfying $0 < |x - a| < \delta$.

true. From B) above we want to say that no such desired $\delta$ exists. This means that the statement C) needs to be false for every number $\delta > 0$. So we now need to negate the statement C). This is easy:

D) $|f(x) - L| \geq \epsilon$ for at least one $x$ satisfying $0 < | x - a| < \delta$.

So we arrive finally at the complete negation of A) by combining B) and D):

E) $L \neq \lim_{x \to a}f(x)$ if for some particular $\epsilon > 0$ and every $\delta > 0$ we have $|f(x) - L| \geq \epsilon$ for at least one value of $x$ satisfying $0 < |x - a| < \delta$.

Now we know how to ensure that the test fails when $L = 10, f(x) = x + 2, a = 1$. What we need is to find one particular value of $\epsilon > 0$ as stated in E) above. Now $|f(x) - L| = |x - 8|$ and our intuitive argument is that $x$ is near $1$ so that $|x - 8|$ is near $7$ and hence we just need to take any $\epsilon < 7$.
Let us choose $\epsilon = 5$ and it is easy to show that E) holds. Note that $|x - 8| \geq 5$ if $x \leq 3$. For any $\delta > 0$ we just need to choose $x$ in the interval $(1, k)$ where $k = \min(1 + \delta, 3)$. This will ensure that $0 < |x - 1| < \delta$ and also $1 < x < 3$. Clearly then $|x - 8| > 5$ for any such $x$.
Note: In order to apply the limit test in case of passing we have to find one $\delta > 0$ for each $\epsilon > 0$, whereas for failure we just need to find one suitable $\epsilon$. So the definition of limit is far easier to apply in cases of failure. What this implies is that the definition of limit is not very helpful in finding the limit and hence we need theorems on "algebra of limits" and certain "standard limit formulas" to evaluate limits.
Further Note: The conceptual understanding of limit is simply an exercise in basic (really basic not even going upto $AM \geq GM$) inequalities and a reasonable level of logic (understanding of quantifiers "for all" $\forall$ and "there exists" $\exists$). Thus it is far simpler than dealing with complicated algebraic/trigonometric manipulation. It is an irony that students find it difficult compared to school algebra courses.
A: The $\epsilon$-$\delta$ definition of limit is:
A limit, $\lim\limits_{x\to c} f(x)$, is said to exist and to equal $L$ if and only if the following statement is true:

For all $\epsilon>0$, there exists some $\delta>0$ such that the implication $|x-c|<\delta\Rightarrow |f(x)-L|<\epsilon$ is true for all $x$.

The negation of this yields the following:
The limit $\lim_{x\to c}$ either doesn't exist or is not equal to $L$ if and only if the following statement is true:

There exists some $\epsilon>0$ such that regardless which $\delta>0$ you pick, you can find some $x$ such that both $|x-c|<\delta$ and $|f(x)-L|\geq \epsilon$ 

That is not to say that every $\epsilon$ you can do this and it is not to say that every $x$ within that range will cause a problem, but at least for one $\epsilon$ and at least one $x$ within that range.  Taking large values of $\epsilon$ as you have, you can still find appropriate $\delta$'s, but if you were to take a smaller $\epsilon$ then you will find strange things happening.
Suppose $\epsilon=1$.  Then you have $|(x+2)-10|<1$ which implies that $|x-8|<1$ which implies that $-1<(x-8)<1$ which finally implies that $7<x<9$.  However, note that regardless which $\delta$ you pick, you can pick $x=\min\{1+0.9\delta,2\}$ to get either $|x-1|<2\leq \delta$ implying that $1<x<3$ or that $|x-1|<\delta\leq 2$, again implying that $1<x<3$, but this is in direct contradiction with our choice of $\epsilon$ which should have meant that $7<x<9$.
