Are the following two statements about limit true or false? and why? 
*

*If $\exists a $, $b$, $\forall \varepsilon >0$, $|a-b| <\varepsilon $，then $a=b$

*If $\forall \varepsilon >0, \exists a,b,$ $|a-b| <\varepsilon$,   then $a=b$
I know the second is true, but I don't know what's wrong with the first one and the discrepancy between them!
 A: The first seems correct. 
Fix $a,b$. For any given $\epsilon$, if $|a-b|<\epsilon$, then $a=b$.
You are given two numbers and you are told that the difference between the two numbers can be arbitrarily small, then a has be equal to b. You need not write $\exists$ at the first place.
The second is wrong:
Fix $\epsilon=1$, let $a=0.5$, $b=0$. Then $|0.5-1|<1$ but $0.5\neq 1$.
A: I think that also the second statement is false. Maybe you should write:
 "if $a$ and $b$ are such that $|a-b|< \epsilon$ for all $\epsilon>0$ then $a=b$".
A: The notation of you formulas is not clear to me.
The syntax of the language is something like


*

*$\forall$ variable(predicate)  

*$\exists$ variable(predicate)

*(predicate) logical_operator (predicate)

*$\lnot$ (predicate)

*...


There are of course a lot of shortcuts like operator precedence. Another is 
$\forall \varepsilon >0$ (predicate)
means
$\forall \varepsilon( (\varepsilon>0) \implies$ (predicate))
or
$\exists \varepsilon >0$ (predicate)
means
$\exists \varepsilon( (\varepsilon>0) \land$ (predicate))
But we avoid shortcuts because I don't exactly know them.
So let's investigate your sentence.
Does your first one mean
$$(\exists a (\exists b( \forall \varepsilon((\varepsilon >0)\implies (\mid a-b \mid <\varepsilon ))))) \implies (a=b)$$
I think that does not make sense. $a$ and $b$ are bounded variables on the left side and free on the right side. So it's meaning is the same as
$$(\exists x (\exists y( \forall \varepsilon((\varepsilon >0)\implies (\mid x-y \mid <\varepsilon ))))) \implies (a=b)$$
Does this make sense? What is $a$ and $b$?
Or do you mean
$$\exists a (\exists b( \forall \varepsilon((\varepsilon >0) \implies  (\mid a-b \mid <\varepsilon )) \implies (a=b)))$$
Then all variables are bound but the meaning is, there exist some special values $a$ and $b$ with this property. This is true for example $a=b=1$ (or $a=b=\text{anything else}$. But I think that is not what you want to say.
There are some other possible notation when one changes the parantheses but I think none of them is the one you want. Also you can introduce parantheses in the second expression but I think you will not get what  you want.
I think you want to say:
$$\forall a ( \forall b ( (\forall \varepsilon ((\varepsilon>0)\implies(\mid a-b \mid <\varepsilon))) \implies (a=b)))  $$
This means: If you have two arbitrary number $a$ and $b$, then if for all positive $\varepsilon$ the value $\mid a-b \mid <\varepsilon$ then $a$ and $b$ must be equal.
