0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – J.T. Mar 27 '18 at 20:36
  • $\begingroup$ Use the $\epsilon$-$\delta$ definition of the limit with this value of $\delta$ and see where it gets you. $\endgroup$ – Rob Arthan Mar 27 '18 at 21:10
  • $\begingroup$ 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. $\endgroup$ – J.T. Mar 27 '18 at 22:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.