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Work So Far: I know the general definition of $\varepsilon-\delta$ continuity at a point $x_0$: $\forall \varepsilon > 0$ $\exists \delta > 0$ such that

$$|x-x_0| < \delta \implies |f(x)-f(x_0)| < \varepsilon$$

Of course, as usual, it is best to begin by setting out to find the necessary $\delta$. The starting point for my problem is thus

$$|x-1| < \delta \implies \left| \frac{1}{x+2} - \frac 1 3 \right| < \varepsilon$$

Through successive manipulations, we see that

\begin{align*} \left| \frac{1}{x+2} - \frac 1 3 \right| &= \left| \frac{3}{3(x+2)} - \frac{x+2}{3(x+2)} \right| \\ &= \left| \frac{3-(x+2)}{3(x+2)} \right| \\ &= \left| \frac{-x+1}{3(x+2)} \right| \\ &= \frac{|x-1|}{3|x+2|} \\ &< \varepsilon \end{align*}

Clearly the assumption $|x-1| < \delta$ is meant to be used here; however, $\delta$ shouldn't be a function of $x$, so I need to do something with the $|x+2|$ in the denominator...

The Question: How do I proceed from here? (Preferably using different functions as examples because this is a homework problem and I want to ultimately do it myself.)

From what I understand, I imagine this is a case where $\delta$ should be chosen to be the minimum of a pair of values, one (usually) a constant, and the other to be a multiple of $\varepsilon$.

However when I've seen such problems in the past - e.g. as examples in texts, or as MSE posts when searching for an answer to this - what to use for $\delta$ has often felt like it's just been "pulled out of thin air," so to speak. A thorough explanation as to how one derives those values (again, in particular alongside examples) would be greatly appreciated, because this is something that's been bugging me for a long time and I've never had a proper answer.

And thanks for the insights you can offer!

Théophile
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PrincessEev
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1 Answers1

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Since we're interested in $x\to 1$, then we can find a way to replace $|x+2|$ by a convenient constant. If $|x-1| < 1$, for example, then $|x+2| > 2$.

Using this fact, $$\frac{|x-1|}{3|x+2|} < \frac{|x-1|}6.$$

Now, if $|x-1| < \delta$, then we would have $$\frac{|x-1|}{3|x+2|} < \frac\delta6.$$

Wouldn't it be convenient if $\frac\delta6$ were equal to $\varepsilon$? In other words, if $\delta=6\varepsilon$?

So this is where you get to rewrite your proof, pulling $\delta = 6\varepsilon$ seemingly out of thin air...

Théophile
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  • Looks like the real issue was me making a typo then lol. (I was working everything in terms of $|x-2|$ and getting confused as to reach anything meaningful from it. Didn't think to check my own work!) I suppose I still need to figure out how to handle cases where we choose, say,

    $$\delta := \min(\delta_1,\delta_2)$$

    or whatever, but I suppose that can be shelved for a while since this other post did help somewhat there.

     – PrincessEev Sep 27 '20 at 00:46
  • Or, wait. Is it the fact that we assume $\delta < 1$ that lets us choose

    $$\delta := \min \left( 6 \varepsilon, 1 \right)$$

    or am I getting something confused?

     – PrincessEev Sep 27 '20 at 00:59
  • @EeveeTrainer Oh, right, I meant to bring up the typo when I edited your question, but I forgot. And yes, you're right to clean up the proof by taking $\delta = \min (6\varepsilon, 1)$. I had glossed over that (I covered the case where $|x-1|<1$ — but what about the alternative?) to get to the main point of where the "magic numbers" in these kinds of proofs come from. The reason everything falls into place so neatly is that someone has already worked out what they want to happen, then has gone back and picked the right numbers accordingly. – Théophile Sep 27 '20 at 02:58
  • So indeed, since it's convenient to assume here that $|x-1| < 1$, then by choosing $\delta = \min(6\varepsilon, 1)$, that covers everything; we don't need to worry about the case $|x-1| \ge 1$. – Théophile Sep 27 '20 at 03:00