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My teacher proposed this question as a challenge proof to do on our own and I can't seem to get it. Was Wondering if anyone could give me a hint or help me on the process to completing it.

Let $a \in \mathbb{R}$. Let $f$ be a function defined, at least, on an interval centred at $a$, except possibly at $a$. Let $L \in \mathbb{R}$.

If $\lim\limits_{ x→a-} f(x) = \lim\limits_{ x→a+} f(x) = L$ then $\lim\limits_{ x→a} f(x) $ = $L$.

We are supposed to prove this using the "delta-epsilon" proof style. Thanks in advance!!

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  • $\begingroup$ If we now add $\lim\limits_{ x→a-} f(x) = \lim\limits_{ x→a+} f(x) = L$ $\impliedby$ $\lim\limits_{ x→a} f(x) $ = $L$. We can turn this into an if and only if proof. I think the other direction is rather easy, since $\exists \delta$ that lets us approach this limit from both sides, both limits exist and we can easily write out the proof by following the definition. Your direction is often considered to be the hard part, this direction follows almost instantly $\endgroup$ – Wesley Strik Dec 10 '18 at 11:29
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Start by writing your hypothesis in terms of delta-epsilon, and what you want to prove, that is

$lim_{x\to a^{-}}f(x)=lim_{x\to a^{+}}f(x)=L$

Implies that given $\epsilon >0$ there exist $\delta_1,\delta_2 >0$ such that if:

$x \in (a-\delta_1,a)$ then $|f(x)-f(a)| < \epsilon$ (which is the formal definition of $lim_{x\to a^{-}}f(x)=L$)

and if

$x \in (a,a+\delta_2)$ then $|f(x)-f(a)| < \epsilon$ (which is the formal definition of $lim_{x\to a^{+}}f(x)=L$)

You want to find a $\delta >0$ such that for any $x \in (a-\delta,a+\delta)$ you have that $|f(x)-f(a)|<\epsilon$

What $\delta>0$ are you going to choose to ensure that the last statement is true?

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    $\begingroup$ would it be the minimum of the 2 deltas? $\endgroup$ – Gbravo Oct 17 '15 at 19:15
  • $\begingroup$ Exactly! That one will work! $\endgroup$ – Joaquin Liniado Oct 17 '15 at 19:16
  • $\begingroup$ alright I think I understand the process, thx!! $\endgroup$ – Gbravo Oct 17 '15 at 19:17
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According to assumptions you have:

$$\eqalign{ & \mathop {\lim }\limits_{x \to {a^ + }} f(x) = L \cr & \exists L:\left( {\forall \varepsilon > 0,\exists {\delta _1} > 0:\left( {\forall x,0 < x - a < {\delta _1} \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right) \cr & \cr & \mathop {\lim }\limits_{x \to {a^ - }} f(x) = L \cr & \exists L:\left( {\forall \varepsilon > 0,\exists {\delta _2} > 0:\left( {\forall x,0 < a - x < {\delta _2} \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right) \cr} $$

Now just combine the two above relations by considering $\delta = \min \left\{ {{\delta _1},{\delta _2}} \right\}$ to get

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

which means

$$\mathop {\lim }\limits_{x \to a} f(x) = L$$

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  • $\begingroup$ is that all we need to show to prove it though? $\endgroup$ – Gbravo Oct 17 '15 at 19:12
  • $\begingroup$ It may need some edits! :) $\endgroup$ – H. R. Oct 17 '15 at 19:16
  • $\begingroup$ yeah, I appreciate the help anyways $\endgroup$ – Gbravo Oct 17 '15 at 19:19
  • $\begingroup$ I added the edits. :) $\endgroup$ – H. R. Oct 17 '15 at 19:20
  • $\begingroup$ Ok yea I see it now, thanks $\endgroup$ – Gbravo Oct 17 '15 at 19:22

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