Comparing absolute value of variables We have
$$0 < |x-a| < \delta $$
How do we get
$$a - \delta < x < a + \delta $$
?
When I add $a$ to both sides I get:
$$a<|x|<a+\delta$$
 A: $|y| < \delta$ iff $-\delta < y < \delta$.
Then you have $-\delta < x-a < \delta$.
Add $a$ throughout to get $a-\delta < x < a+\delta$.
Note that the last equation is not equivalent to the first condition given, for equivalence
you need to add the condition $x \neq a$.
Elaboration of first line:
Suppose $|y| < \delta$. Then either $y \ge 0$, in which case $0 \le y < \delta$ (and so $-\delta < y < \delta$) or
$y < 0$ in which case $|y| = -y $ and we have $0 \le -y < \delta$. Multiplying
the latter by $-1$ (and reversing the inequalities) gives
$-\delta < y \le 0$ (and so $-\delta < y < \delta$).
Now suppose $-\delta < y < \delta$. If $y \ge 0$, we have $|y|=y < \delta$.
If $y < 0$ we have $|y|=-y$ and since $y > -\delta$ we have (multiplying across by $-1$) $-y < -\delta$ and so $|y|=-y < \delta$.
A: By definition,
$|x|=x$ if $x\ge0$ and $|x|=-x$ if $x<0$. Thus, $|x|\ge0$ for all $x$.
Here,
$$|x-a|<\delta$$

Case 1: $x-a>0$
$$|x-a|=x-a<\delta$$
$$x<a+\delta$$
As $\delta>0$, we have $-\delta<0$, i.e., $x-a>-\delta$ which gives $x>a-\delta$.

Case 2: $x-a<0$
$$|x-a|=-(x-a)<\delta$$
$$x-a>-\delta$$
$$x>a-\delta$$
As $\delta>0$, we have $x-a<\delta$ which gives $x<a+\delta$.
Thus,
$$a-\delta<x<a+\delta$$
A: It doesn't actually give that result. For instance, $x = a$ satisfies the second conclusion, but not the first. 
A: One has $0\lt |b|\lt c\iff -c\lt b\lt c$. We apply that to $|x-a|\lt\delta$ to get
$$-\delta\lt x-a\lt \delta$$
And this implies
$$a-\delta\lt x\lt a+\delta$$
A: The other answers are correct, but may not help you in the future.
I think you should begin with an example and some words, rather than trying to "add $a$ to both sides".
Think about the situation when (for example) $a = 3$ and $\delta = 1$. Then the condition says you care about the values of $x$ for which the distance from $3$ is less than $1$ but bigger than $0$. You should draw the picture that shows why  $x$ is between $3-1=2$ and $3+1=4$ (values $1$ away from $3$), but not $3$ itself.
