Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/194



When, for a particular value of the independent variable, a function takes on one of the forms

$$\begin{smallmatrix}\frac{0}{0}, \frac{\infty}{\infty},\ 0 \cdot \infty,\ \infty - \infty,\ 0^0,\ \infty^0,\ 1^{\infty},\end{smallmatrix}$$

it is said to be indeterminate, and the function is not defined for that value of the independent variable by the given analytical expression.

For example, suppose we have

$$\begin{smallmatrix}y = \frac{f(x)}{F(x)},\end{smallmatrix}$$

where for some value of the variable, as x = a,

$$\begin{smallmatrix}f(a) = 0,\ F(a) = 0.\end{smallmatrix}$$

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, §18) 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.

If when x = a the function $$f(x)$$ assumes an indeterminate form, then

$$\begin{smallmatrix}\lim_{x = a} f(x)\end{smallmatrix}$$

is taken as the value of $$f(x)$$ for x = a.

The assumption of this limiting value makes $$f(x)$$ continuous for x = a. This agrees with the theorem under Case II, p. 15, and also with our practice in Chapter III, where several functions assuming the indeterminate form $$\begin{smallmatrix}\frac{0}{0}\end{smallmatrix}$$ were evaluated. Thus, for x = 2 the function $$\begin{smallmatrix}\frac{x^2 - 4}{x - 2}\end{smallmatrix}$$ assumes the form $$\begin{smallmatrix}\frac{0}{0}\end{smallmatrix}$$ but

$$\begin{smallmatrix}\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4.\end{smallmatrix}$$

Hence 4 is taken as the value of the function for x = 2. Let us now illustrate graphically the fact that if we assume 4 as the value of the function for x = 2, then the function is continuous for x = 2.