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:

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.
\begin{figure}\mbox{\subfigure[removable]{\epsfig{file=DiscontLine.eps,height=4c...
...d
\subfigure[removable]{\epsfig{file=xLnAbsx.eps,height=4cm}
}
}\end{figure}

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

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

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

Figure 3: A removable discontinuity.
\begin{figure}\mbox{\epsfig{file=xTimesSin1_x.eps,height=6cm}}\end{figure}

Noah Dana-Picard 2007-12-28