# Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/195

Placing each factor separately equal to zero, we have

x = 2, and y = x + 2.

the equation of the line CD. Also, on CD, when x = 2, we get

y = MP = 4.

In plotting, the loci of these equations are found to be the two lines AB and CD respectively. Since there are infinitely many points on the line AB having the abscissa 2, it is clear that when x = 2 (= OM), the value of y (or the function) may be taken as any number whatever; but when x is different from 2, it is seen from the graph of the function that the corresponding value of y (or the function) is always found from

y = x + 2,

which we saw was also the limiting value of y (or the function) for x = 2; and it is evident from geometrical considerations that if we assume 4 as the value of the function for x = 2, then the function is continuous for x = 2.

Similarly, several of the examples given in Chapter III illustrate how the limiting values of many functions assuming indeterminate forms may be found by employing suitable algebraic or trigonometric transformations, and how in general these limiting values make the corresponding functions continuous at the points in question. The most general methods, however, for evaluating indeterminate forms depend on differentiation.

111. Evaluation of the indeterminate form $\tfrac{0}{0}$. Given a function of the form $\tfrac{f(x)}{F(x)}$ such that $f(a) = 0$ and $F(a) = 0$; that is, the function takes on the indeterminate form $\tfrac{0}{0}$ when a is substituted for x. It is then required to find

$\lim_{x \to a} \frac{f(x)}{F(x)}.$

Draw the graphs of the functions $f(x)$ and $F(x)$. Since, by hypothesis, $f(a) = 0$ and $F(a)$ = 0, these graphs intersect at (a, 0).