Definition of limit and axiom of choice In the definition of limit of a function ($\epsilon-\delta$ definition) we say certain statements such as for every $\epsilon>0$ there exist $\delta>0$ ....
Now my question is, is a choice function required to ensure that for every $\epsilon$ there exist a $\delta>0$? Moreover, how we can we test whether a mathematical statement depends on axiom of choice or not?
I really don't know whether this question has any meaning or not, so please help me.

The question doesn't make much sense, and it seems that you are misunderstanding the definition.
We define continuity by $\varepsilon$-$\delta$ at a point $x$ in such way. If you want to show that a function is continuous at $x$ it suffices to show that every $\varepsilon$ has such $\delta$. We are not required to assign in a single stroke all the $\varepsilon$'s with fitting $\delta$'s. We just have to show that given $\varepsilon>0$ there is some $\delta$ which satisfies the statement.
On the other hand, the axiom of choice is required in order to show that continuity by $\varepsilon$-$\delta$ at a point is equivalent to continuity using sequences (i.e. $f(x)=\lim_{x_n\to x_0}f(x_n)$ for every $x_n\to x$). The axiom of countable choice suffices here.
For more details on that last part, see the entire discussion on: Continuity and the Axiom of Choice
Finally, to prove that the axiom of choice is required to prove $\varphi$ we can either show that $\varphi$ implies something we already know to require the axiom of choice (or a fragment thereof), or that we can construct a model of $\sf ZF$ in which $\varphi$ fails (of course it makes sense only if $\sf ZFC$ proves $\varphi$ in the first place).
For more on that, you can read:


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*How do we know we need the axiom of choice for some theorem?

*We cannot write this function
