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Here is a proof for $\lim_{x\to2} 3x^2 = 12$


We are given some $\epsilon > 0$, and we need to find $\delta$ such that

$0 < |x-2| < \delta \Rightarrow |3x^2 - 12| < \epsilon$

The inequality $|3x^2 - 12| < \epsilon$ will be more useful if it is in terms of $x-2$ rather than x, since the inequality $0 < |x-2| < \delta$ is in terms of $x-2$. For simplicity, let $z = x-2$. Then we wish to find $\delta$ such that

$0 < |z| < \delta \Rightarrow |3(z+2)^2 - 12| < \epsilon$

We can simplify this to

$0 < |z| < \delta \Rightarrow |3z^2 + 12z| < \epsilon$

However, we know that $|3z^2 +12z|\leq|3z^2|+|12z|=3z^2 + 12|z|$. So it suffices to find $\delta$ such that

$0 < |z| < \delta \Rightarrow 3z^2 + 12|z| < \epsilon$

If $ 0 < |z| < \delta$, then $3z^2 + 12|z| < 3\delta^2 + 12\delta = 3\delta(4 + \delta)$. Thus it suffices to choose $\delta$ such that

$3\delta(4 + \delta) < \epsilon$

The $4 + \epsilon$ term is somewhat annoying. We can make it simpler by assuming that $\delta \leq 1$.

If we assume that $\delta \leq 1$, then $4+\delta \leq 5$, and the inequality that we need becomes simply

$3\delta(4+\delta) \leq 15\delta < \epsilon$

To force this to be true, we select $\delta = \frac{\epsilon}{15}$. (In the unlikely event that $\epsilon > 15$, we can just take $\delta = 1$.) We then conclude that

$0 < |z| < \delta \Rightarrow |3x^2 - 12| < 3\delta(4 + \delta) \leq 15\delta=\epsilon$.

Thus, for any $\epsilon < 15$, we have found that $\delta = \frac{\epsilon}{15}$ satisfies the $\delta$-$\epsilon$ condition:

$0 < |x-2| < \delta \Rightarrow |3x^2 - 12| < \epsilon$

and hence we have stablished that $\lim_{x\to2} 3x^2 = 12$


What I don't understand is why just because something is greater than $|3z^2 + 12z|$ It suffices to find $\delta$ using it. I mean by that logic couldn't I say something like

$|3z^2 + 12z| \leq |3z^2 + 12z| + z^{5000000}$, so it suffices to find $\delta$ such that $0 < |z| < \delta \Rightarrow|3z^2 + 12z| + z^{5000000} < \epsilon$

Any help would be appreciated

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  • $\begingroup$ To make one's life easier, one can show $\lim_{x\to 2}x^2=4$ instead. $\endgroup$ – Jack Jul 16 '17 at 20:06
  • $\begingroup$ sorry I copied the proof directly from my textbook to show the part I didn't understand $\endgroup$ – Bhaskar Jul 16 '17 at 20:08
  • $\begingroup$ That would be correct but it would not make the problem easier. Replacing the $4+\delta$ by 5 made the problem easier. $\endgroup$ – Ian Jul 16 '17 at 20:08
  • $\begingroup$ but I still don't understand why you can do that. Especially the parts I bolded. $\endgroup$ – Bhaskar Jul 16 '17 at 20:11
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    $\begingroup$ It's not the same, but the latter $\delta$ works for the former case. If you want $u(x) < \epsilon$, but instead make $v(x) < \epsilon$ where $u(x) < v(x)$, then you also get $u(x) < \epsilon$. $\endgroup$ – md2perpe Jul 16 '17 at 21:43
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[This is an answer directly answering your confusion.]

... We can simplify this to $$0 < |z| < \delta \Rightarrow |3z^2 + 12z| < \epsilon\tag{a}$$

However, we know that $|3z^2 +12z|\leq|3z^2|+|12z|=3z^2 + 12|z|$. So it suffices to find $\delta$ such that $$0 < |z| < \delta \Rightarrow 3z^2 + 12|z| < \epsilon\tag{b}$$

So here is your question: why "it suffices to find $\delta$ such that (b) is true" while what we want is (a).

Suppose you have found a $\delta$ such that (b) is true. Then since "we know that" $$ |3z^2 +12z|\leq|3z^2|+|12z|=3z^2 + 12|z|\tag{c}, $$ we have $$ 0 < |z| < \delta \Rightarrow |3z^2 + 12z| \leq 3z^2 + 12|z|< \epsilon. $$


The reason to work on (b) instead of (a) is not only it is logically correct but also it is useful to find out a $\delta$ one needs. It is logically correct but useless to say that

it suffices to find $\delta$ such that $$ 0<|z|<\delta\Rightarrow |3z^2+12z|+z^{10000}<\epsilon. $$

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[Not a direct answer to your question. But too long to be a comment.] I appreciate your effort writing down the long question. But I strongly dislike the way your textbook gives the proof: it makes things look so complicated and that's not the way we do analysis in practice!


Here is what one could do. Let $0<\epsilon<1$. One wants to find $\delta>0$ so that the following implication is true $$ 0<|x-2|<\delta\Rightarrow |x^2-4|<\epsilon. $$ Observe that $|x^2-4|<\epsilon$ is equivalent to $$ |x-2|\cdot|x+2|<\epsilon\tag{1} $$ One can see that when $x$ is getting "close" to $2$, $|x-2|$ can be as small as possible while $|x+2|$ remains bounded by some fixed number. This is the essential point to give the proof. To make it precise, choose $\delta=\epsilon$. Then if $0<|x-2|<\delta$, we have $$ |x-2|\cdot |x+2|\leqslant\epsilon (\epsilon +4)<5\epsilon\tag{2} $$ where we use the triangle inequality $|x+2|\leqslant |x-2|+4$. I claim that (2) completes the proof by the following easy exercise.

Exercise. Show that the following statements are equivalent:

  • There exists some constant $C$ such that for every $\epsilon>0$, $|A|\leqslant C\epsilon$.

  • For every $\epsilon>0$, $|A|\leqslant \epsilon$.

  • For every $\epsilon$ with $0<\epsilon<1$, $|A|\leqslant \epsilon$.


I would like to repeat a remark I made in another answer:

One important tactic that is seldom mentioned in elementary real analysis or calculus textbooks is that when doing an estimate in analysis, one should never worry about the constant in front of one's epsilon. As Terry Tao points out in one of his excellent blog posts on problem solving strategies in real analysis:

Don’t worry too much about exactly what $\varepsilon$ (or $\delta$, or $N$, etc.) needs to be. It can usually be chosen or tweaked later if necessary.

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  • $\begingroup$ I don't completely understand how you reached (2). Where did you get $\epsilon(\epsilon + 4)$ and $5\epsilon$ $\endgroup$ – Bhaskar Jul 16 '17 at 20:44
  • $\begingroup$ Note in the beginning that I set $0<\epsilon<1$. $\endgroup$ – Jack Jul 16 '17 at 20:47
  • $\begingroup$ Oh ok I get it. $\endgroup$ – Bhaskar Jul 16 '17 at 20:50
  • $\begingroup$ But still out of curiosity, do you have any way to explain the answer to my original question. It's frustrating to me, because I see that used in so many other proofs, and I don't understand why it works. $\endgroup$ – Bhaskar Jul 16 '17 at 20:52
  • $\begingroup$ Strictly, you are not allowed to set limits on $\epsilon$ (like $\epsilon<1$). However, if $\epsilon>1$ and you select $\delta$ so that $|x^2-4|<1$ then obviously $|x^2-4|<\epsilon$ in that case. $\endgroup$ – md2perpe Jul 16 '17 at 21:37
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Probably your confusion is due to dealing with two concepts that can seem (but are not) contradictory: these are, given a function,

  1. its majorant that is greater than or equal to the function pointwisely
  2. the subdomain of majorant where it is minorized that is smaller (in the sense of inclusion) than or equal to the correspondig subdomain of the function.

In your case the function is: $$f(z)=\lvert3z^2+12z\rvert$$ its subdomain where it is minorized is: $$F_{\varepsilon}=\{z:f(z)<\varepsilon\}$$ The majorant is: $$g(z)=3z^2+12\lvert z\rvert$$ with the corresponding minorizing subdomain: $$G_{\varepsilon}=\{z:g(z)<\varepsilon\}$$ Now being:

$$f(z)\le g(z)$$ it is (and here is probably your confusion; think deeply on it and then doubts will disapper forever) $$F_{\varepsilon}\supseteq G_{\varepsilon}\tag{1}$$

Now if correspondingly to a given $\varepsilon$ there is a $\delta_{\varepsilon}$ for which the set: $$C_{\delta_{\varepsilon}}=\{z:0<\lvert z\rvert<\delta_{\varepsilon}\}$$ is such that $$C_{\delta_{\varepsilon}}\subseteq G_{\varepsilon}$$ then, a fortiori for $(1)$, it is also $$C_{\delta_{\varepsilon}}\subseteq F_{\varepsilon}$$ This is the condition needed by the definition of the limit (that has to be checked for all $\varepsilon>0$).


More schematically, you want to know why \begin{align*} (\forall z \textrm{ such that }0<\lvert z\rvert <\delta_\varepsilon &\implies \lvert3z^2+12z\rvert<\varepsilon) \\&\impliedby \tag{is implied by}\\ (\forall z \textrm{ such that }0<\lvert z\rvert <\delta_\varepsilon &\implies 3z^2+12\lvert z\rvert<\varepsilon) \end{align*}

Proof \begin{align*} (\forall z \textrm{ such that }0<\lvert z\rvert <\delta_\varepsilon &\implies \lvert3z^2+12z\rvert<\varepsilon) \\&\iff \tag{1}\\ (\{z:0<\lvert z\rvert <\delta_\varepsilon \} &\subseteq\{z:\lvert3z^2+12z\rvert<\varepsilon\}) \\ &\impliedby \tag{2}\\ (\{z:0<\lvert z\rvert <\delta_\varepsilon \} &\subseteq\{z:3z^2+12\lvert z\rvert<\varepsilon\}) \\ &\iff \tag{3}\\ (\forall z \textrm{ such that }0<\lvert z\rvert <\delta_\varepsilon &\implies 3z^2+12\lvert z\rvert<\varepsilon) \end{align*}

where

$(1)$ and $(3)$ is for the known relation between implication and subset whereby $$(\forall x \textrm{ such that }P(x)\implies Q(x))\iff\left(\{x:P(x)\}\subseteq\{x:Q(x)\}\right)$$

$(2)$ is because $\forall z \lvert3z^2+12z\rvert\le3z^2+12\lvert z\rvert$ and therefore $$\{z:\lvert3z^2+12z\rvert<\varepsilon\}\supseteq\{z:3z^2+12\lvert z\rvert<\varepsilon\}$$

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The wonderful thing about these $\varepsilon - \delta$ problems is that you can 'clean up your tracks' and make it look like you know what you are doing with a math 'word processor'.

There are many ways to attack these problems and there is no cookie cutter approach. If you have problems understanding a demonstration, try solving it yourself on your own terms, then go back to recheck the other approach.

I want to show some "sketch-it-out" work and then give a finished answer; it can be instructive to see more than one technique for the same problem.


Sketch

Simple algebra shows,
$3x^2 - 12 = (x-2) (12 + 3(x-2)) = 12(x-2) + 3 (x-2)^2$
If $K \gt 0$, then
$|12(x-2)| < K/2 \text{ iff } |(x-2)| \lt K/24$
AND
$| 3 (x-2)^2| < K/2 \text{ iff } |(x-2)| \lt \sqrt {K/6}$
Also, if $K \lt 1$, $K/24 \lt \sqrt {K/6}$


Formal Proof

$\lim_{x\to2} 3x^2 = 12$

Let the $\varepsilon \gt 0$ 'challenge' be given; we can harmlessly assume that the challenge is less than $1$, so that

$\frac{3 \varepsilon^2}{24^2} \lt \varepsilon/2$.

Set $\delta = \frac{\varepsilon}{24}$ and take any $x$ with $|x-2| \lt \delta$.

Since $3x^2 - 12 = 12(x-2) + 3 (x-2)^2$, by the triangle inequality,

$\tag 1 |3x^2 - 12| \le 12|(x-2)| + 3 |(x-2)|^2$

Both of the terms on the RHS of (1) are less than $\epsilon/2$, so the limit does indeed converge to $12$.

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