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.

I don't get your first question, could you elaborate? As for the second one, a mathematical statement can't "depend" on the axiom of choice, a proof of this statement can.
Now, if your question was "does a proof of the statement require the axiom of choice", then this is very difficult to answer. More formally, for a formula $\phi$, you are asking whether $\mathrm{ZF} \not\vdash \phi$ but $\mathrm{ZFC} \vdash \phi$ is true. To illustrate how this is difficult, take $\phi$ to be the countable axiom of choice. Then it has not been known for a long time whether $\mathrm{ZF} \not\vdash \phi$ was true, altough $\mathrm{ZFC}\vdash \phi$ is trivial.
