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I want to prove $\lim_{x \to 2} \frac{3}{x+1} = 1 $ using the $ \delta $-$\epsilon$ definition. How can I select a $\delta $ based on $ \epsilon $ so this function stays bounded within 1?

Digitallis
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user124820929
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    Hello! Have you ever seen an $\epsilon$-$\delta$ proof involving a limit of a function of the form $\frac{a}{bx + c}$ before? Did you feel like you understood it? What was the problem when you tried to recreate this proof in this situation? If you add details like this, your question will likely be better received! – Izaak van Dongen Sep 28 '23 at 18:57
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    Please edit your title so that it describes the question. There are hundreds of epsilon delta questions (hence the tag). – Digitallis Sep 28 '23 at 18:58
  • No, I've never seen an example of this form. I've only seen an example involving a linear function. – user124820929 Sep 28 '23 at 19:01
  • So you actually want to prove continuity. The limit is otherwise trivial because the function is perfectly well defined and continuous at x=2. – Al.G. Sep 28 '23 at 19:07
  • In that case I admire your bravery! In future it could also help you to try and look for such a proof somewhere else in case that inspires you (I would certainly expect any text about real analysis to mention this at some point.) There is some good discussion here and here on this site that you may find helpful! They are hard to find if you don't know to search for "continuity". (In case you don't know - "$1/x$ is continuous at $c$" means "$\lim_{x \to c} 1/x = 1/c$", so they're very similar!) – Izaak van Dongen Sep 28 '23 at 19:59
  • Is the solution below a correct approach? – user124820929 Sep 28 '23 at 20:06
  • Quotiesnts in an $~\epsilon, ~\delta ~$ limit problem normally require special handling. $0 < |x - 2| < \delta \implies $ $$x \neq 2 ~~\text{and} ~~2 - \delta < x < 2 + \delta \implies 3 - \delta < x + 1 < 3 + \delta.$$ Now, I customarily impose an additional artificial upper bound on $~\delta,~$ such as $~\delta \leq (1/10).~$ ...see next comment – user2661923 Sep 28 '23 at 21:12
  • @user124820929, It's not correct, but it's not too far off. But it can be fixed quite easily by instead choosing $\delta = \min(\epsilon, 1)$, for example. Then whenever $|x - 2| < \delta$, we have $|2 - x| < \epsilon$ and also $1/|x + 1| < 1$, so the result follows. This is the usual approach with such functions. In this case if you pick $\delta = \epsilon$, the only "problem" happens if $\epsilon$ is large. But you'll come to understand that such problems can always be avoided, since all we really care about is the very very small $\epsilon$s. – Izaak van Dongen Sep 28 '23 at 21:18
  • Then, for example: $$\frac{1}{3 - \delta} = (1/3) + \frac{\delta/3}{3 - \delta} \leq (1/3) + \frac{\delta/3}{3 - .1} < (1/3) + [(.4) \times (\delta/3)].$$ Similarly, $$\frac{1}{3 - \delta} = (1/3) + \frac{\delta/3}{3 - \delta} > (1/3) + \frac{\delta/3}{3 - 0} = (1/3) + \delta/9.$$ – user2661923 Sep 28 '23 at 21:20
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    @IzaakvanDongen I agree, imposing an additional constraint on $~\delta~$ is the way to go, per my previous comments. I usually opt for $~\delta \leq (1/10).$ – user2661923 Sep 28 '23 at 21:21
  • @user2661923, fair enough :). I think in this case a $\delta$ as big as $\min(2, \varepsilon)$ would even still work! But I do love a good erring on the side of caution! I've written many limit proofs involving things like $N + 100$, and many set theory proofs involving things like $V_{\lambda + 10}$. OP - the key point to take away is that it doesn't matter so much what constant $C$ you use when you ask for $\delta = \min(C, \epsilon)$, so long as it's "reasonably small". In analysis it's often fine to just pick a somewhat arbitrary small/large number in these situations! – Izaak van Dongen Sep 28 '23 at 21:30

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Since the question is how to select a $\delta$ and not which delta works, I'll try to answer that, so it's more like a tutorial and won't give you the particular answer explicitly so that you find it yourself.

The problem is to prove that at $x=2$, $$\forall\varepsilon>0\exists\delta>0\forall x':\left|x'-x\right|\to\left|f(x')-f(x)\right|\le\varepsilon$$ where $f(x')=3/(x'+1)$. (I'll consider $x$ "fixed" and $x'$ "floating" in my answer)

  1. Let $\varepsilon>0$ be given; now we have to find some $\delta>0$ such that (plugging $x=2$ and $f(x)$): $$\forall x':\left|x'-2\right|\leq\delta\to\left|\frac{3}{x'+1}-\frac{3}{2+1}\right|\le\varepsilon,$$ or more simply, $\forall x':\left|x'-2\right|\leq\delta\to\left|\frac{3}{x'+1}-1\right|\le\varepsilon$. The natural thing to do when we don't see a direct solution is to expand the definitions. Our target inequality is $\left|\frac{3}{x'+1}-1\right|\le\varepsilon$, and by definition it means $-\varepsilon\le\frac{3}{x'+1}-1\le\varepsilon$.

  2. Now stop and think (so far we only did blind simplifications and definition expansions). We want the target equality fulfilled for all $x'$ sufficiently close to $x=2$, i.e. whenever $|x'-2|\le\delta$. So let's try to answer the question, for which $x'$ is it fulfilled? For which $x'$ does $-\varepsilon\le\frac{3}{x'+1}-1\le\varepsilon$? Solving the inequality for $x'$ we get $$\frac{3}{1+\varepsilon}-1\le x'\le\frac{3}{1-\varepsilon}-1.$$

  1. Now stop and think again. The question is, what $\delta$ makes $|x'-2|\le\delta$ imply the target equality? We want $x'$ clamped in the interval $\left[\frac{3}{1+\varepsilon}-1,\frac{3}{1-\varepsilon}-1\right]$. And the question is to find an interval $[2-\delta,2+\delta]$ (corresponding to the $|x'-2|\le\delta$ condition) which fits within the target interval. Now that's a purely arithmetic question that I believe you'll be able to solve. And interestingly, it will give you also the best delta of all!
Al.G.
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Since $ lim (x) = 2 => |x-2| < 1 $ $\\$ => in this restricted domain, $ x $ belongs to $ (1, 3) $

Let's choose $\delta = 2\epsilon$.

$$ 0 < | x - 2| < \delta = 2\epsilon $$ $ $ $ => \left| \frac{3}{x+1} - 1 \right| = \left| \frac{2-x}{x+1} \right| = \frac{|x - 2|}{|x+1|}< \frac{|x-2|}{2} $ (since max($\frac{1}{|x+1|}) = \frac12$ in the above domain)

Finally, as $| x - 2| <\delta = 2\epsilon$, $$\left| \frac{3}{x+1} - 1 \right| < \epsilon$$

This method of restricting the domain can be used for any limit proof of the form $\frac{a}{bx+c}$.

Devansh Agarwal
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  • I ended up choosing $\delta$ = min{1/2, $\epsilon$/2}. Say $\delta$=1. Then 1/|x+1| is less than 1/2.

    So $\left| \frac{3}{x+1} - 1 \right|$ < 1/2 * 2$\epsilon$. Is this correct?

     – user124820929 Sep 28 '23 at 22:37
  • $\delta = \min{1/2,\epsilon/2}$ is consistent with the $\delta$ that Al.G. found -- that is, your $\delta$ is always smaller, so it works at least as well as the other one. But you lost me at "Say $\delta = 1$." How can you say $\delta = 1$ when you just defined it so it can never be greater than $1/2$? – David K Sep 29 '23 at 05:31