Removable discontinuity.

Definition 5.3.1   Take a function $f$ defined on a pointed neighborhood $V'$ of $x_0$ . Suppose that $f$ has a finite limit $a$ at $x_0$ . Then we say that $f$ has a removable discontinuity at $x_0$ .

Define a function as follows:

• If $x \in V', \; g(x)=f(x)$ ;
• $g(x_0)=a$ .

Then $g$ is continuous at $x_0$ .

Example 5.3.2       

1. The discontinuity at 0 of $f: x \mapsto \frac 1x$ is not removable, as $f$ has two infinite one-sided limits at 0 (see Figure 2(d)).
2. The discontinuity at 1 of $f: x \mapsto \frac {x^2-1}{x-1}$ is removable, as $f$ has a finite limit, equal to 2, at 1 (see Figure 2(e)).
3. Let $f: \; x \mapsto 1+ x \ln \vert x\vert$ , defined over $\mathbb{R}^*$ . We can easily prove that $f$ has a limit at 0 and that this limit is equal to 1. Thus 0 is a removable discontinuity of $f$ .

Figure 2: Removable (or not) discontinuity.

Example 5.3.3   Take $f(x)=x \sin \frac 1x$ for $x \neq 0$ . Look for a limit of $f$ at 0: $f$ is the product of a bounded function (as the sine function is bounded over $\mathbb{R}$ , we have $-1 \leq \sin 1/x \leq 1$ ) and a function whose limit at 0 is equal to 0. Therefore, by Theorem thm sandwich, we have

$$\underset{x \rightarrow 0}{\lim} f(x)=0.$$

This means that $f$ has a removable discontinuity at 0.

Figure 3: A removable discontinuity.