Proof of continuity at a point

Question

I'm struggling to understand the method of this proof. It's as follows:

Prove that $f(x) = x^2$ is continuous at $x = 2$.

Proof: We have $$|f(x) - f(2)| = |x^2 - 4| = |x+2||x-2|$$

If we insist that $|x-2| < 1$, then $1<x<3$ and so $|x+2| < 5$. Thus, given $\epsilon > 0$, if we choose $\delta =$ min$\{1, \epsilon / 5\}$, then

$$|f(x) - f(2)| = |x+2||x-2| < 5|x-2| < \epsilon$$

whenever $|x-2| < \delta$, proving that $f$ is continuous at 2.

I understand the application of the definition for the most part. My confusion is coming from the part in boldface, where the author "insists" that $|x-2| < 1$. Why can the author do that? The natural domain of $f$ is all of $\mathbb{R}$, so why can we hone in on just $1<x<3$?

Answer 1

While I think Bernard's answer is nice and succinct, I think that it is more appropriate for someone who is a bit more familiar with the definitions and concepts. I would like to answer the question in terms of the definition of continuity at a point.

Given $\epsilon>0$, we want to find a $\delta>0$ such that whenever $|x-2|<\delta$, then $|f(x)-f(2)|<\epsilon$. Well, since we are searching for a $\delta>0$, we may as well assume that $\delta < 1$, and try to find an appropriate $\delta$ that is less than $1$.

Answer 2

That is because continuity is a local notion: to show that a function is continuous at a point $x_0$ of its domain, it is enough to prove that its restriction to an open neighbourhood of $x_0$ (contained in its domain) is continuous.

Answer 3

The author has put it backwards (but it's still valid). We are free to choose $\delta$ however we like, as long as it does the job; the author chooses $\delta=\min\{1,\varepsilon/5\}$, and it is this choice (with $\delta\le 1$) that ensures $|x-2|< 1$ whenever $|x-2|<\delta$.