Prove that $\lim_{x \to 2} x^2 = 4$.

Question

Prove that $\lim_{x \to 2} x^2 = 4$.

Let $\varepsilon >0$. Let $\delta=\min\{1,\frac \varepsilon 5\}$.

Suppose 0 < $|x-2|$ < $\delta$ $\rightarrow 1<x<3$, since $\delta =1$. (#)

$|x^2-4|=|x+2||x-2|<\delta|x+2|<5\delta=\varepsilon$ (since $\delta = \frac \varepsilon 5$). (*)

This is the proof for that limit. But, I am confused with $\min$ function. When my lecturer proved this, he use $\delta=1$ for (#), and $\delta = \frac \varepsilon 5$ for (*). Can we arbitrarily select one value in $\min$ function? if we can, why is that??

Answer 1

This is too long for a comment, and I don't know whether it actually answers your question, so here goes:

It is rather common practice to limit the size of $\delta$ in proofs like these. That's because the function $f(x) = x^2$, as you get further and further away from $2$, gets steeper and steeper. This implies that single expression for a $\delta$ which works for all $\varepsilon$ (particularily the "optimal" $\delta$, i.e. the largest $\delta$ possible) is going to involve $\sqrt\varepsilon$ and stuff.

However, the definition of limits doesn't require us to find an optimal $\delta$, it requires us to find any $\delta$ which works. And any $\delta$ smaller than the optimal one works. Besides, continuity and limit at $x = 2$ aren't really about what happens for large $\varepsilon$'s and $\delta$'s, and $x$'s far away from $2$. It's about what happens for small $\varepsilon$'s and $\delta$'s, and for $x$'s close to $2$.

Thus declaring that "whatever happens, I will never choose my $\delta$ to be larger than $1$" means that I can ignore what the function does for $x\leq 1$ or $x\geq 3$. That won't be relevant, since we have made a choice which forces $(2-\delta, 2+\delta)\subseteq (1, 3)$. Therefore it doesn't matter that the function gets steeper and steeper further out. On $(1, 3)$ there is a limit to how steep the function gets, and that allows us to use a nice expression like $\frac\varepsilon5$ instead.

So setting $\delta = \min(1, \frac\varepsilon5)$ is saying is that we use $\delta = 1$ for all $\varepsilon$ we can, and when $\varepsilon$ gets too small for $\delta = 1$ to work, we set $\delta = \frac\varepsilon5$, and then the meat of the proof is showing that it works in this case too (and, of course, there is some work involved in finding $\frac\varepsilon5$ in the first place).

Answer 2

That is almost correct, but not quite. But simply writing “since $\delta\leqslant1$” instead of “since $\delta=1$” and writing “since $\delta\leqslant\frac\varepsilon5$” instead of “since $\delta=\frac\varepsilon5$” will make it just fine.