Epsilon-Delta proof of $\lim_{x\to 2} x^2=4$

Question

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!

Answer 1

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.

Answer 2

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 $\delta=\frac{1}{2}\min \left\{1,\frac{\epsilon}{5}\right\}$.