Proof of continuity for $f(x) = x^{-1}$

Question

Consider this proof:

This is from my professors notes. I understand her steps and all but I cant seem to figure out why $\delta$ is bounded to $\frac{|a|}{2}$ instead of 1. Would bounding $\delta$ to 1 work in this case? Any help is appreciated.

Answer 1

According to the definition of continuity, we need to find, for all $\varepsilon > 0$, a $\delta > 0$ (depending on $\varepsilon$) such that $$|x - a| < \delta \implies |f(x) - f(a)| < \varepsilon.$$ Note, the condition $|x - a| < \delta$ is equivalent to saying $x$ lies on the interval $(a - \delta, a + \delta)$. By choosing and enforcing a limit on $\delta$, we gain the power to eliminate problematic values of $x$ from this interval.

In our case, the problematic value of $x$ is $x = 0$. At $x = 0$, the function is undefined, but worse, around $x = 0$, the function grows arbitrarily large. We cannot possibly expect $|f(x) - f(a)| < \varepsilon$ for very large values of $f(x)$. So, it is a good idea to impose a limit on $\delta$, not only so that $(a - \delta, a + \delta)$ does not contain $0$, but also that it contains no points around $0$.

This is what limiting $\delta \le \frac{|a|}{2}$ does. This is half the distance from $a$ to $0$, which means $(x - a, x + a)$ will only stretch half the distance towards $0$. The proof bears out that this helps us. By limiting $\delta \le \frac{|a|}{2}$, we can conclude $|x| > \frac{|a|}{2} > 0$ from $|x - a| < \delta$. This means $\frac{1}{|x|} < \frac{2}{|a|}$, which puts a lid on how big the $\frac{1}{|x|}$ factor can be.

Will limiting $\delta \le 1$ do the same job? Not really. If $a$ lies between $-1$ and $1$, and we only limit $\delta \le 1$, then the interval $(a - \delta, a + \delta)$ could include $0$ (depending on $a$ and $\delta$), which means there will be nearby points with large function values. It means you have problematic points in your interval.

For example, if $a = 1/2$, and I assume $\delta \le 1$, then I can conclude that $|x - 1/2| < 1$, i.e. $-1/2 < x < 3/2$. On this interval, the function $f$ is unbounded! We should have limited $\delta$ to be less than some number strictly less than $1/2$, so that $0$ (and points around $0$) are excluded. This will make $f(x)$ at least abounded when $|x - 1/2| < \delta$.