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!


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.

  • $\begingroup$ Ok, the concern about the quare roots is fair. However, if you can find the optimal bounds, would that also be a valid method? $\endgroup$ – GambitSquared Jun 30 '15 at 13:22
  • $\begingroup$ @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. $\endgroup$ – Ian Jun 30 '15 at 13:25
  • $\begingroup$ Thanks you Ian! $\endgroup$ – GambitSquared Jun 30 '15 at 13:26
  • $\begingroup$ Where does the 1 in $\min\{1, \epsilon/5\}$ come from? $\endgroup$ – Björn Lindqvist Aug 29 '17 at 15:54
  • $\begingroup$ @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 $|x+2|$. $\endgroup$ – Ian Aug 29 '17 at 16:04

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 $|x-2|<\delta$.


$|x^2-4|=|x-2||x+2|$ 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 $\delta=\frac{1}{2}\min \left\{1,\frac{\epsilon}{5}\right\}$.

  • $\begingroup$ 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 $3^2-4=5$). $\endgroup$ – Ian Jun 30 '15 at 14:44
  • $\begingroup$ "this is the step that only works in the reals" As opposed to what? $\endgroup$ – Did Jul 13 '15 at 15:31
  • $\begingroup$ The complex numbers. In the complex numbers is |z|<a then the statement -a<z<a is false unless Im(z)=0 (real). $\endgroup$ – Aleksandar Jul 13 '15 at 16:28
  • $\begingroup$ 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 @.) $\endgroup$ – Did Jul 19 '15 at 17:49
  • $\begingroup$ (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. $\endgroup$ – Did Jul 22 '15 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.