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!
Best Answer
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.