.

Other singularities require more analysis to diagnose. Consider the functions $ \frac{\sin x}{x}$, $ \frac{\sin x}{\vert x\vert}$ and $ \frac{\sin x}{1 - \cos x}$ at the point $ x = 0$. All three functions evaluate to $ \frac {0}{0}$ at that point, but have different kinds of singularities. The first has a removable discontinuity, the second has a finite discontinuity and the third has an infinite discontinuity. See Figure 13.5.

Figure: The functions $ \frac{\sin x}{x}$, $ \frac{\sin x}{\vert x\vert}$ and $ \frac{\sin x}{1 - \cos x}$.
\includegraphics[height=4cm,width=10cm]{disc3.eps}

An expression that evaluates (for a particular value of the independent variable) to $ \frac {0}{0}$, $ \frac {\infty }{\infty }$, $ 0 \cdot \infty $, $ \infty -\infty $, $ 1^\infty$, $ 0^0$ or $ \infty ^0$ is called an indeterminate. A function $ h(x)$ which takes an indeterminate form at $ x = a$ is not defined for $ x = a$ by the given analytical expression. For example, suppose we have

$\displaystyle y = \frac{f(x)}{g(x)},
$

where for some value of the variable, as

$\displaystyle f(a) = 0,\ \ \ \ g(a) = 0.
$

For this value of $ x$ our function is not defined and we may therefore assign to it any value we please. It is evident from what has gone before (Case II, [§3.6]) that it is desirable to assign to the function a value that will make it continuous when $ x = a$ whenever it is possible to do so.

david joyner 2008-11-22