# How to prove $\lim_{x \to 2} x^2 = 4$ and make sure it's correct?

I would like to specify that english is not my primary language so I had to do some translation here, I do really hope every mathematical term I used is correct, if not please forgive me :)

Let's say I want to prove that $\lim_{x \to 2} x^2 = 4$, from what I've read, in order to do that, the solution to $|f(x)-l|<\epsilon$ must contain a neighborhood of $x_{0}$ such as $x_{0}-\delta<x<x_{0}+\delta$ which proves that $|x-x_{0}|<\delta$

So in this case $|x^2-4|<\epsilon$ must contain a neighborhood of $2$, so I proceed by solving that:

$-\epsilon<x^2-4<\epsilon$

$4-\epsilon<x^2<4+\epsilon$

$\sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}$

and this is as far as my book goes

The limit is proven true because $\sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}$ is a neighborhood of $x_{0}$ which is 2, so when $x\to2$ then $y\to4$

but what I don't understand is: how do we know that $\sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}$ is really a neighborhood of $2$? That doesn't seem obvious to me.

Let's say that instead of a "true" limit I wanted to prove (Or prove wrong) a "false" limit such as $\lim_{x \to 2} x^2 = 8$ which is not true. How do I do that? I can do the usual steps:

$-\epsilon<x^2-8<\epsilon$

$8-\epsilon<x^2<8+\epsilon$

$\sqrt{8-\epsilon}<x<\sqrt{8+\epsilon}$

but how do I check if this is really a neighborhood of 2 so that $|x-x_{0}|<\delta$ or not? How do I know if this limit is really proven or not?

PS: This is what I've learned reading my book, if anything I said is wrong, please let me know! And if you know a faster/better/simpler method of proving a limit please tell me!

• Unless the book worked out examples other than this, this is a crappy book. Nobody proves things like that. Aug 12, 2017 at 15:22
• @Cauchy translated word for word: "To prove that $\lim_{x\to x_{0}}=l$ we need to solve $|f(x)-l|<\epsilon$ and prove that the solutions contains a neighborhood of $x_{0}$ for every $\epsilon>0$" Aug 12, 2017 at 15:28
• Maybe of interest: math.stackexchange.com/questions/195726/… Aug 12, 2017 at 17:30

I suppose by a neighborhood of $2$ you are talking about an open interval of center $2$. If so, then note that $0 < \varepsilon < 4$ implies that $\sqrt{4-\varepsilon} < 2 < \sqrt{4+\varepsilon}$. Let $a := \min \{ 2-\sqrt{4-\varepsilon}, \sqrt{4+\varepsilon}-2 \}$. Then $x \in (2-a,2+a)$ implies $|x^{2}-4| < \varepsilon$. Here the function $x \mapsto x^{2}$ is defined at $x=2$, so there is no need to exclude the possibility $x=2$. All in all, this is what the author tries to convey, I think. I am silent on whether the author succeeds though...

To see why this argument works, note that it shows that for every $0 < \varepsilon < 4$ we can construct some $\delta > 0$ (the $a$ above is an instance) such that $|x-2| < \delta$ implies $|x^{2}-4| < \varepsilon$. For every $\varepsilon \geq 4$, we use the same $\delta$, which suffices.

I am not in any conflict of interest with the book or the author or anything related to it :). I would like to say that the proof is an intention to make things easy. It is a subject matter if it is good or not. I would say that the choice of the method is with pedagogical consideration but the lack of explanation backfires.

You can do it with triangle inequality. Note that $|x^{2}-4| = |x-2||x+2|$ for all real $x$. If $|x-2| < 1$, then $|x| - 2 \leq |x-2| < 1$ by triangle inequality, so $|x| < 3$ and hence $|x+2| \leq |x| + 2 < 5$ by triangle inequality, so $|x-2||x+2| < 5|x-2|$. If $\varepsilon > 0$, then $5|x-2| < \varepsilon$ if $|x-2| < \varepsilon/5$. Combining all the majorants for $|x-2|$, we conclude that $|x-2| < \min \{ 1, \varepsilon/5 \}$ implies $|x^{2}-4| < \varepsilon$.

• People are saying that this method of proving limits is bad, do you know where I can find a better one? Aug 12, 2017 at 15:44
• @EliaPerantoni, Hold on, I got a minute :). Aug 12, 2017 at 15:45
• @EliaPerantoni, You may check my newest version now. Aug 12, 2017 at 15:50
• Thansk! It will take me some time to understand this since I don't know what triangle inequalities are but I'll try :) Aug 12, 2017 at 15:53
• But as I've said in a comment, can't I just assume that $\epsilon$ is extremely small yet positive and see what happens? In this case we would have $\sqrt{4-0^+}<x<\sqrt{4+0^+}$ becomes $2^-<x<2^+$ ? I've tried this with different excersises and it seems to work pretty well! Aug 12, 2017 at 15:59

The square root function is increasing and continuous and $\sqrt{4} = 2$, hence for $\varepsilon>0$ we have that $\sqrt{4-\varepsilon}<2$ and $\sqrt{4+\varepsilon}>2$.

• It is better to write ".......is increasing and continuous AT THE NEIGHBORHOOD OF $x=2$ and ......." Aug 12, 2017 at 15:34
• Because $\epsilon$ can as small as we want as long as it is $\epsilon>0$ right? So if we choose an incredibly small $\epsilon$ so $\sqrt{4-\epsilon}$ and $\sqrt{4+\epsilon}$ are almost equal to $\sqrt{4}$, now we have $2^-<x<2^+$ so we can assume $x_{0}=2$ and the limit is proven? Is this correct? Aug 12, 2017 at 15:37
• @EliaPerantoni Yes, that's basically correct. (I say "basically" because your intuition is correct, however it depends on how rigorously you want to prove it).
– Eff
Aug 12, 2017 at 18:07