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I have seen and understand the delta-epsilon proof of the limit of $x^2$ for $x\to2$, such as explained here: https://www.youtube.com/watch?v=gLpQgWWXgMM

Now I am wondering, is there also another way? How about this:

Verify that $\lim x^2=4$ (for $x\to2$)

STEP A: Express epsilon in terms of $x$: \begin{align} |x^2-4| &< \varepsilon\\ -\varepsilon &< x^2-4 < \varepsilon\\ 4-\varepsilon &< x^2 < 4+\varepsilon\\ \sqrt{4-\varepsilon} &< x < \sqrt{4+\varepsilon} \end{align}

STEP B: Express delta in terms of $x$ \begin{align} |x-2| &< \delta\\ -\delta &< x-2 < \delta\\ 2-\delta &< x < 2+\delta \end{align}

STEP C: Now we can express $\delta$ in terms of $\varepsilon$ hence proving the limit.
If we take $\delta=\min\{-2+\sqrt{4+\varepsilon},2-\sqrt{4-\varepsilon}\}$ then the limit is proven

Did I make any mistake? Thanks! Cheers!

GambitSquared
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2 Answers2

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That seems a little bit confusing to me. For one thing there is a concern about the square roots being defined. I would recommend working as follows.

Note that $|x^2-4|=|x+2||x-2|$. Looking at this, we can make the factor of $|x-2|$ small, but we have to be sure that the factor of $|x+2|$ is not too big. But that's not so bad: if $|x-2|<1$ then $x \in (1,3)$, and so $|x+2|<5$. Hence $|x^2-4|<5|x-2|$. So if $|x-2|$ is also less than $\frac{\varepsilon}{5}$ then $|x^2-4|<\varepsilon$.

So now you're done: given $\varepsilon > 0$, choose $\delta=\min \{ 1,\varepsilon/5 \}$. With this choice, if $|x-2|<\delta$ then $|x^2-4|<\varepsilon$.

This kind of approach will work in a lot more problems than your kind of approach, because usually you can't find the optimal bounds like you tried to do. Instead you usually make a series of estimates, which are hopefully accurate enough to get you what you need.

Ian
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  • Ok, the concern about the quare roots is fair. However, if you can find the optimal bounds, would that also be a valid method? – GambitSquared Jun 30 '15 at 13:22
  • @ImreVégh While it technically is, it's harder when it's possible and it's quite frequently impossible. It's good to practice with getting suboptimal bounds which are still adequate to prove the result at hand. – Ian Jun 30 '15 at 13:25
  • Thanks you Ian! – GambitSquared Jun 30 '15 at 13:26
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    Where does the 1 in $\min{1, \epsilon/5}$ come from? – Gaslight Deceive Subvert Aug 29 '17 at 15:54
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    @BjörnLindqvist It comes from the $1$ in $|x-2|<1$ from the previous paragraph. But that choice was arbitrary. I could've made it $2$, then $|x-2|<2$ gives $|x|<4$ so $|x^2-4|<6|x-2|$, so $\delta=\min { 2,\epsilon/6 }$ works too. I just need some cutoff so I have a bound for $|x+2|$. – Ian Aug 29 '17 at 16:04
  • Thank you! I wish profs demonstrating this proof would use a truly arbitrary number like 47, because a 1 looks chosen for a specific reason. :) – Gaslight Deceive Subvert Aug 30 '17 at 13:43
  • @BjörnLindqvist I think a better choice if you want to emphasize that it is arbitrary would be some weird "small" number like $1/17$ or something. That emphasizes the "locality" of the calculation. – Ian Aug 30 '17 at 14:00
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This is the general solution for a problem like this but the final steps only work in $\mathbb{R}$.

Problem: $\forall\epsilon>0$ $\exists\delta>0$ such that $|x-2|<\delta$ implies $|x^2-4|<\epsilon$

This may seem a bit over-complicated but it is saying, $$\lim_{x \to 2} x^2=4$$ if you can derive $|x^2-4|<\epsilon$ from $|x-2|<\delta$.

Solution:

$|x^2-4|=|x-2||x+2|$ therefore $|x-2|<\frac{\epsilon}{|x+2|}$. We can't have that since delta must be a function of just epsilon. The theorem only applies when $x$ is close to 2 we can set, $|x-2|<1$ (this is the step that only works in the reals) this implies $-1<x-2<1$ which implies $x<3$.

Therefore, $\frac{\epsilon}{|x+2|}<\frac{\epsilon}{5}$ therefore the $\delta=\min \left\{1,\frac{\epsilon}{5}\right\}$.

Post Script: I believe we can go so far as to set $\delta=\frac{1}{2}\min \left\{1,\frac{\epsilon}{5}\right\}$.

Aleksandar
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  • You could get a sharper bound with the mean value theorem. The result would be $\delta=\min { a,\frac{\varepsilon}{4+b} }$ where $b$ is a positive number which goes to zero as $a$ goes to zero. That's because the derivative of $x^2$ at $2$ is $4$. When $a=1$, the best possible $b$ is also $1$ (since $3^2-4=5$). – Ian Jun 30 '15 at 14:44
  • "this is the step that only works in the reals" As opposed to what? – Did Jul 13 '15 at 15:31
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    The complex numbers. In the complex numbers is |z|<a then the statement -a<z<a is false unless Im(z)=0 (real). – Aleksandar Jul 13 '15 at 16:28
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    Yeah sure, but plenty of other steps also require working with real numbers, not complex hence to draw the attention on this specific step might be misleading. (Unrelated: Please use @.) – Did Jul 19 '15 at 17:49
  • (Comment of archival value.) I previously posted here three comments, addressed to the OP and describing in details three serious defects of the paragraph "Solution" of their answer. Together with other comments of no value wisely deleted by a mod, these comments went away, perhaps no so wisely. But frankly, seeing the other answer by @Ian, which is most adequate, I feel no urge to repost these three comments here. – Did Jul 22 '15 at 07:46
  • Why not | x- 2 |<0.1 or | x − 2 |< 0.01, etc? Why it has to be 1? – Carlitos_30 Jun 05 '23 at 23:39
  • I'm arriving terribly late, but for archival purposes: this solution is horribly wrong! Going from worst to, well less-worst.... "the theorem only applies when $x$ is close to 2" is incorrect: the theorem applies for every $\varepsilon > 0$ (not to mention the aspect of: what is "close to" 2?? So, you cannot state $|x-2| < 1$ as a consequence of the definition of limit. Next: the derivation of $\cdots < \epsilon/5$ is also wrong: $x<3$ does not imply that $\cdots < \epsilon/5$ ($x = 0$ satisfies the condition $x<3$, and yet it results in $\epsilon/2$, which is not less than $\epsilon/5$) – Cal-linux May 14 '24 at 10:52
  • Next (ran out of characters in the previous comment), the very initial statement is wrong: I know it is only the phrasing, but this is a mathematical proof, so being rigurous with both the formulas and the sentences is crucial. The quoted definition is not the same as saying that lim....=4 if you can derive ...... (because the part after "you can derive" makes no sense, since you're not saying anything about $\varepsilon$ or $\delta$). – Cal-linux May 14 '24 at 11:00
  • And lastly, being a bit more picky/pedantic: the comment of "only works in the reals", in addition to being entirely unnecessary and for that matter distracting, it is also wrong: it does not only work in the reals: it also works in the rationals, in the integers, etc. Regardless, the main point is: why do you need to specify that as part of this proof/derivation? who said anything about complex values that would make us explicitly say that whatever step only works in R? (again, with the "only" being incorrect) – Cal-linux May 14 '24 at 11:04