I'm being asked to prove that $\lim_{x \rightarrow c}f(x)=\lim_{x \rightarrow c}f\mid_J(x)$, given that $J \subseteq I \subseteq \mathbb{R}$ are open intervals, $c \in J$, and $f: I - \left \{ c \right \} \rightarrow \mathbb{R}$ is a function. This question has been asked before, but was vague and not answered.
Showing that $\lim_{x \rightarrow c}f(x)= L \Rightarrow \lim_{x \rightarrow c}f\mid_J(x) = L$ is simple enough using the $\varepsilon-\delta$ definition and the fact that $J \subseteq I$. Unfortunately, I've hit a stumbling block with the other direction.
If $\lim_{x \rightarrow c}f\mid_J(x) = L$, then I know that for each $\varepsilon > 0$, there will exist some $\delta > 0$ such that $\left\vert x-c \right\vert < \delta$ and $\left\vert\ f(x)-L\ \right\vert < \varepsilon$, but only when $x \in J$. The feedback I was given indicates that I should use the fact that $J$ is open to find a second delta, which doesn't make much sense at the moment.
Given that $J$ is open, I know that there will exist a $\delta$ such that $\left[ x-\delta, x+\delta \right ] \subseteq J \subseteq I$, but I'm not clear on whether or not that's a different delta, or whether or not that has anything to do with what I need.
There's also the "bounding" stuff, where (possibly):
$$x \in J - \left \{ c \right \} \cup \left( x-\delta, x+\delta \right ) \subseteq I - \left \{ c \right \} \cup \left( x-\delta, x+\delta \right )$$
but I'm not convinced that has anything to do with anything either.
Hint: you have two deltas satisfying two different useful properties, so give them different names, say $\delta_1$ and $\delta_2$. If you let $\delta = \min(\delta_1, \delta_2)$, what can you deduce?