# Given $J \subseteq I \subseteq \mathbb{R}$ open intervals, prove $\lim_{x \rightarrow c}f(x)=\lim_{x \rightarrow c}f\mid_J(x)$

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.

Let without loss of generality (why?) $\epsilon$ be so little as that $(x-\epsilon, x+\epsilon)$ is contained in the smaller interval $J$ then what do you know about about $f(x)$ when $x$ enters this environment of c?
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
• Okay, so I can use the limit for $\delta_1$ and the open interval for $\delta_2$. If I suppose $x \in I - \left\{c\right\}$, setting $\delta = \mathrm{min}(\delta_1,\delta_2)$ gives $\left\vert x-c \right\vert < \delta_2 \Rightarrow -\delta_2 < x-c < \delta_2 \Rightarrow c -\delta_2 < x < c + \delta_2$. Then, presumably, because $c \in J$, it follows somehow that the limit stays the same for $x \in I$. Is that about the right track? – J.T. Mar 27 '18 at 20:36
• Use the $\epsilon$-$\delta$ definition of the limit with this value of $\delta$ and see where it gets you. – Rob Arthan Mar 27 '18 at 21:10
• That's what I was trying to do... I think I should be able to derive $\left\vert\ f(x)-L\ \right\vert < \varepsilon$ from all of that, but I can't see how it happens. Pardon my hard-headedness, but my brain has a hard time with $\varepsilon-\delta$ stuff. – J.T. Mar 27 '18 at 22:37