Discrete Math Logic HomeworkConsider the following statement: ∀ ϵ > 0 , ∃...

Step 1
The usual definitions of "converse" and "contrapositive" used in logic only apply to implications, which are statements of the form $A\Rightarrow B$. The converse of $A\Rightarrow B$ is $B\Rightarrow A$, and the contrapositive is $(\mathrm{\neg }B)\Rightarrow (\mathrm{\neg }A)$, where $\mathrm{\neg }A$ is the negation of A.
Step 2
Because the statement you wrote has two quantifiers at the front, it is not an implication, and the usual definitions do not apply to it. Therefore, you should ask your instructor, or consult your notes, to learn what the instructor wants you to do. It will be difficult to find much help in the usual textbooks or reference sources because these do not give any definition for the "converse" or "contrapositive" of statements that are not implications.

Step 1
$\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:(|x-a|<\delta ⟹|f(x)-L|<ϵ).$
Step 2
Perhaps what you are asked to do is to replace the quantified (internal) implication with it's contrapositive:
$\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:(|x-a|<\delta \to |f(x)-L|<ϵ)$
$⟺\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:[\mathrm{\neg }(|f(x)-L|<ϵ)\to \mathrm{\neg }(|x-a|<\delta )]$
$⟺\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:[\mathrm{\neg }\mathrm{\neg }(|f(x)-L|<ϵ)\vee \mathrm{\neg }(|x-a|<\delta )]$
$⟺\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:\mathrm{\neg }[\mathrm{\neg }(|f(x)-L|<ϵ)\wedge (|x-a|<\delta )]$
$⟺\mathrm{\neg }\mathrm{\exists }ϵ>0,\mathrm{\forall }\delta >0:[(|f(x)-L|\ge ϵ)\wedge (|x-a|<\delta )]$