Elements of the Differential and Integral Calculus/Chapter XIII

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Elements of the Differential and Integral Calculus by William Anthony Granville Chapter XIII, § 105–115

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CHAPTER XIII

THEOREM OF MEAN VALUE. INDETERMINATE FORMS

105. Rolle's Theorem. Let y = f(x) be a continuous single-valued function of x, vanishing for x = a and x = b, and suppose that f'(x) changes continuously when x varies from a to b. The function will then be represented graphically by a continuous curve as in the figure. Geometric intuition shows us at once that for at least one value of x between a and b the tangent is parallel to the axis of X (as at P); that is, the slope is zero. This illustrates Rolle's Theorem:

If f(x) vanishes when x = a and x = b, and f(x) and f'(x) are continuous for all values of x from x = a to x = b, then f'(x) will be zero for at least one value of x between a and b.

This theorem is obviously true, because as x increases from a to b, f(x) cannot always increase or always decrease as x increases, since f(a) = 0 and f(b) = 0. Hence for at least one value of x between a and b, f(x) must cease to increase and begin to decrease, or else cease to decrease and begin to increase; and for that particular value of x the first derivative must be zero (§ 81).

That Rolle's Theorem does not apply when f(x) or f'(x) are discontinuous is illustrated as follows:

Fig. a shows the graph of a function which is discontinuous ($= \infty$) for x = c, a value lying between a and b. Fig. b shows a continuous function whose first derivative is discontinuous ($= \infty$) for such an intermediate value x = c. In either case it is seen that at no point on the graph between x = a and x = b does the tangent (or curve) be,come parallel to OX.

106. The Theorem of Mean Value. Consider the quantity Q defined by the equation

(A)	$\begin{smallmatrix}\frac{f(b) - f(a)}{b - a} = Q,\end{smallmatrix}$ or
(B)	f(b) − f(a) − (b − a)Q = 0.
Let F(x) be a function formed by replacing b by x in the left-hand member of (B); that is,
(C)	F(x) = f(x) − f(a) − (x − a)Q.
From (B),	F(b) = 0, and from (C), F(a) = 0;
therefore, by Rolle's Theorem (§105) F'(x) must be zero for at least one value of x between a and b, say for x1. But by differentiating (C) we get
	F'(x) = f'(x) − Q.
Therefore, since	F'(x1) = 0, then also f'(x1) − Q = 0,
and	Q = f'(x1).
Substituting this value of Q in (A), we get the Theorem of Mean Value,
(44)	$\begin{smallmatrix}\frac{f(b) - f(a)}{b - a} = f'(x_1),\ a < x_1 < b\end{smallmatrix}$

where in general all we know about x₁ is that it lies between a and b.

The Theorem of Mean Value interpreted Geometrically. Let the curve in the figure be the locus of

y = f(x).

Take OC = a and OD = b; then f(a) = CA and f(b) = DB, giving AE = b − a and EB = f(b) − f(a).

Therefore the slope of the chord AB is

(D)	$\begin{smallmatrix}\tan EAB = \frac{EB}{AE} = \frac{f(b) - f(a)}{b - a}.\end{smallmatrix}$
There is at least one point on the curve between A and B (as P) where the tangent (or curve) is parallel to the chord AB. If the abscissa of P is Xl' the slope at P is
(E)	tan t = f'(x1) = tan EAB.

Equating (D) and (E), we get

$\begin{smallmatrix}\frac{f(b) - f(a)}{b - a} = f'(x_1),\end{smallmatrix}$

which is the Theorem of Mean Value.

The student should draw curves (as the one in §105 to show that there may be more than one such point in the interval; and curves to illustrate, on the other hand, that the theorem may not be true if f(x) becomes discontinuous for any value of x between a and b (Fig. a, §105), or if f'(x) becomes discontinuous (Fig. b, §105).

Clearing (44) of fractions, we may also write the theorem in the form

(45)	f(b) = f(a) + (b − a)f'(x1).
Let b = a + Δa; then b − a = Δa, and since x1 is a number lying between a and b, we may write
	$x_l = a + \theta \cdot \Delta a,$
where θ is a positive proper fraction. Substituting in (45), we get another form of the Theorem of Mean Value.
(46)	$\begin{smallmatrix}f(a + \Delta a) - f(a) = \Delta a f'(a + \theta \cdot \Delta a). 0 < \theta < 1\end{smallmatrix}$

107. The Extended Theorem of Mean Value. Following the method of the last section, let R be defined by the equation

(A)	$\begin{smallmatrix}f(b) - f(a) - (b - a)f'(a) - \frac{1}{2} (x - a)^2 = 0.\end{smallmatrix}$
Let F(x) be a function formed by replacing b by x in the left-hand member of (A); that is,
(B)	$\begin{smallmatrix}F(x) = f(x) - f(a) - (x - a) f'(a) - \frac{1}{2} (x - a)^2 R.\end{smallmatrix}$
From (A), F(b) = 0; and from (B), F(a) = 0; therefore, by Rolle's Theorem (§105), at least one value of x between a and b, say x1 will cause F'(X) to vanish. Hence, since
	F'(x) = f'(x) − f'(a) − (x − a)R, we get
	F'(x1) = f'(x1) − f'(a) − (x1 − a)R = 0.
Since F'(x1) = 0 and F'(a) = 0, it is evident that F'(x) also satisfies the conditions of Rolle's Theorem, so that its derivative, namely F''(x), must vanish for at least one value of x between a and x1, say x2, and therefore x2 also lies between a and b. But
	F''(x) = f''(x) − R; therefore F''(x2) = f''(x2) − R = 0,
and	R = f''(x2).

Substituting this result in (A), we get

(C)

$\begin{smallmatrix}f(b) = f(a) + (b - a)f'(a) + \frac{1}{2!} (b - a)^2 f''(x_2).\end{smallmatrix}$

a < x2

In the same manner, if we define S by means of the equation

$\begin{smallmatrix}f(b)\ -\ f(a)\ -\ (b - a) f'(a)\ -\ \frac{1}{2!} (b - a)^2 f''(a)\ -\ \frac{1}{3!}(b - a)^2 f''(a) S\ =\ 0,\end{smallmatrix}$

we can derive the equation

(D)	$f(b)\ =\ f(a)$	$+ (b - a)f'(a) + \frac{1}{2!} (b - a)^2 f''(a)$
		$+ \frac{1}{3!} (b - a)^3 f'''(x_3),$	$a\ < x_3 <\ b$

where x₃ lies between a and b.

By continuing this process we get the general result,

(E)	$f(b)\ =\ f(a)$	$+ \frac{(b - a)}{1!}f'(a) + \frac{(b - a)^2}{2!}f''(a)$
		$+ \frac{(b - a)^3}{3!}f'''(a) + \cdots + \frac{(b - a)^{(n - 1)}}{(n - 1)!}f^{(n - 1)} (a)$
		$+ \frac{(b - a)^n}{n!}f^{(n)} (x_1),$	$a\ < x_1 <\ b$

where x₁ lies between a and b. (E) is called the Extended Theorem of Mean Value.

108. Maxima and minima treated analytically. By making use of the results of the last two sections we can now give a general discussion of maxima and minima of functions of a single independent variable.

Given the function f(x). Let h be a positive number as small as we please; then the definitions given in § 82, may be stated as follows:

If, for all values of x different from a in the interval [a - h, a + h],

(A)	f(x) − f(a) = a negative number,

then f(x) is said to be a maximum when x = a.

If, on the other hand,

(B)	f(x) − f(a) = a positive number,

then f(x) is said to be a minimum when x = a.

Consider the following cases:

I. Let $f'(a) \ne 0.$

From (45), §106, replacing b by x and transposing f(a),

(C)

f(x) − f(a) = (x − a)f'(x1).

a < x1 < x

Since $f'(a) \ne 0$, and f'(x) is assumed as continuous, h may be chosen so small that f'(x) will have the same sign as f'(a) for all values of x in the interval [a - h, a + h]. Therefore f'(x₁) has the same sign as f'(a) (Chap. III). But x - a changes sign according as x is less or greater than a. Therefore, from (C), the difference

f(x) − f(a)

will also change sign, and, by (A) and (B), f(a) will be neither a maximum nor a minimum. This result agrees with the discussion in § 82, where it was shown that for all values of x for which f(x) is a maximum or a minimum, the first derivative f'(x) must vanish.

II. Let f'(a) = 0, and $f''(a) \ne 0$.

From (C), § 107, replacing b by x and transposing f(a),

(D)

$f(x) - f(a) = \frac{(x - a)^2}{2!} f''(x_2).$

a < x2 < x

Since $f''(a) \ne 0$, and f''(x) is assumed as continuous, we may choose our interval [a - h, a + h] so small that f''(x₂) will have the same sign as f''(a) (Chap. III). Also (x − a)² does not change sign. Therefore the second member of (D) will not change sign, and the difference

f(x) − f(a)

will have the same sign for all values of x in the interval [a - h, a + h], and, moreover, this sign will be the same as the sign of f''(a). It therefore follows from our definitions (A) and (B) that

(E)	f(a) is a maximum if f'(a) = 0 and f''(a) = a negative number;
(F)	f(a) is a minimum if f'(a) = 0 and f''(a) = a positive number.

These conditions are the same as (21) and (22), §84.

III. Let f'(a) = f''(a) = 0, and $f'''(a) \ne 0.$

From (D), §107, replacing b by x and transposing f(a),

(G)

$f(x) - f(a) = \frac{1}{3!} (x - a)^3 f'''(x_3).$

a < x3 < x

As before, f'''(x₃) will have the same sign as f'''(a). But (x − a)³ changes its sign from - to + as x increases through a. Therefore the difference

f(x) − f(a)

must change sign, and f(a) is neither a maximum nor a minimum.

IV. Let $f'(a) = f''(a) = \cdots = f^{(n-l)}(a) = 0,$ and $f^{(n)}(a) \ne 0$.
By continuing the process as illustrated in I, II, and III, it is seen that if the first derivative of f(x) which does not vanish for x = a is of even order (= n), then
(47)	f(a) is a maximum if f(n)(a) = a negative number;
(48)	f(a) is a minimum if f(n)(a) = a positive number.
If the first derivative of f(x) which does not vanish for x = a is of odd order, then f(a) will be neither a maximum nor a minimum.

Illustrative Example 1. Examine x³ − 9x² + 24x − 7 for maximum and minimum values.

Solution.	f(x) = x3 − 9x2 + 24x − 7.
	f'(x) = 3x2 − 18x + 24.
Solving	3x2 − 18x + 24 = 0
gives the critical values x = 2 and x = 4. ∴ f'(2) = 0, and f'(4) = 0.
Differentiating again,	f''(x) = 6x − 18.
Since f''(2) = − 6, we know from (47) that f(2) = 13 is a maximum.
Since f''(4) = + 6, we know from (48) that f(4) = 9 is a minimum.

Illustrative Example 2. Examine e^x + 2cos x + e ^{− x} for maximum and minimum values.

Solution.	f(x) = ex + 2cos x + e − x,
	f'(x) = ex − 2sin x − e − x = 0, for x = 0,
	f''(x) = ex − 2cos x + e − x = 0, for x = 0,
	f'''(x) = ex + 2sin x − e − x = 0, for x = 0,
	fiv(x) = ex + 2cos x + e − x = 4, for x = 0.

Hence from (48), f(0) = 4 is a minimum.

EXAMPLES

Examine the following functions for maximum and minimum values, using the method of the last section:

1. 3x4 − 4x3 + 1.	Ans.	x = 1 gives min. = 0;
		x = 0 gives neither.
2. x3 − 6x2 + 12x + 48.		x = 2 gives neither.
3. (x − 1)2(x + 1)3.		x = 1 gives min. = 0 ;
		$\begin{smallmatrix} x = \frac{1}{5} \end{smallmatrix}$ gives max.;
		x = -1 gives neither.

4. Investigate x⁶ − 5x⁴ + 5x³ − 1, at x = 1 and x = 3.

5. Investigate x³ − 3x² + 3x + 7, at x = 1.

6. Show that if the first derivative of f(x) which does not vanish for x = a is of odd order (= n), then f(x) is an increasing or decreasing function when x = a, according as f⁽ⁿ⁾(a) is positive or negative.

109. Indeterminate forms. 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.

110. Evaluation of a function taking on an indeterminate form. 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.

Let	$\begin{smallmatrix}y = \frac{x^2 - 4}{x - 2}\end{smallmatrix}$
This equation may also be written in the form
	y(x − 2)	= (x − 2)(x + 2);
or	(x − 2)(y − x − 2)	= 0.

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).

Applying the Theorem of Mean Value to each of these functions (replacing b by x), we get

f(x) = f(a) + (x − a)f'(x1),	a < xl < x
F(x) = F(a) + (x − a)F'(x2).	a < x2 < x

Since f(a) = 0 and F(a) = 0, we get, after canceling out (x - a),

$\frac{f(x)}{F(x)} = \frac{f'(x_1)}{F'(x_2)}.$

Now let $x \dot= a$; then $x_l \dot= a, x_2 \dot= a$, and

$\lim_{x \to a} f'(x_1) = f'(a), \lim_{x \to a} F'(x_2) = F'(a)$.

(49)

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

$F'(a) \ne 0$

Rule for evaluating the indeterminate form $\tfrac{0}{0}$. Differentiate the numerator for a new numerator and the denominator for a new denominator. The value of this new fraction for the assigned value of the variable will be the limiting value of the original fraction.

In case it so happens that

f'(a) = 0 and F'(a) = 0,

that is, the first derivatives also vanish for x = a, then we still have the indeterminate form $\frac{0}{0}$, and the theorem can be applied anew to the ratio

$\frac{f'(x)}{F'(x)}$

giving us

$\lim_{x \to a} \frac{f(x)}{F(x)} = \frac{f''(a)}{F''(a)}.$ When also f''(a) = 0 and F''(a) = 0, we get in the same manner

$\lim_{x \to a} \frac{f(x)}{F(x)} = \frac{f'''(a)}{F'''(a)},$

and so on.

It may be necessary to repeat this process several times. Illustrative Example 1. Evaluate $\tfrac{f(x)}{F(x)} = \tfrac{x^3 - 3x + 2}{x^3 - x^2 - x - 1}$ when x = 1.

Solution.	$\frac{f(1)}{F(1)} = \left. \frac{x^3 - 3x + 2}{x^3 - x^2 + 1} \right]_{x = 1} = \frac{1 - 3 + 2}{1 - 1 - 1 + 1} = \frac{0}{0}.$ ∴ indeterminate.
	$\frac{f'(1)}{F'(1)} = \left. \frac{3x^2 - 3}{3x^2 - 2x - 1} \right]_{x = 1} = \frac{3 - 3}{3 - 2 - 1} = \frac{0}{0}.$ ∴ indeterminate.
	$\frac{f''(1)}{F''(1)} = \left. \frac{6x}{6x - 2} \right]_{x = 1} = \frac{6}{6 - 2} = \frac{3}{2}.$ Ans.

Illustrative Example 2. Evaluate $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}.$

Solution.	$\frac{f(0)}{F(0} = \left. \frac{e^x - e^{-x} - 2x}{x - \sin x} \right]_{x = 0} = \frac{1 - 1 - 0}{0 - 0} = \frac{0}{0}.$ ∴ indeterminate.
	$\frac{f'(0)}{F'(0)} = \left. \frac{e^x - e^{-x} - 2}{1 - \cos x} \right]_{x = 0} = \frac{1 + 1 - 2}{1 - 1} = \frac{0}{0}.$ ∴ indeterminate.
	$\frac{f''(0)}{F''(0)} = \left. \frac{e^x - e^{-x}}{\sin x} \right]_{x = 0} = \frac{1 - 1}{0} = \frac{0}{0}.$ ∴ indeterminate.
	$\frac{f'''(0)}{F'''(0)} = \left. \frac{e^x - e^{-x}}{\cos x} \right]_{x = 0} = \frac{1 + 1}{1} = 2.$ Ans.

EXAMPLES

Evaluate the following by differentiation:

1. $\lim_{x \to 4} \frac{x^2 - 16}{x^2 + x - 20}.$	Ans.	$\frac{8}{9}$.	9. $\lim_{\theta \to 0} \frac{\theta - \arcsin \theta}{\sin^3 \theta}.$	Ans.	$-\frac{1}{6}.$
2. $\lim_{x \to 1} \frac{x - 1}{x^n - 1}.$		$\frac{1}{n}.$	10. $\lim_{x \to \phi} \frac{\sin x - \sin \phi}{x - \phi}.$		cos ϕ
3. $\lim_{x \to 1} \frac{\log x}{x - 1}$		1.	11. $\lim_{y \to 0} \frac{e^y + \sin y - 1}{\log(1 + y)}.$		2.
4. $\lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x}.$		2.	12. $\lim_{\theta \to 0} \frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1}.$		1.
5. $\lim_{\frac{\tan x - x}{x - \sin x}}.$		2.	13. $\lim_{\phi \to \frac{\pi}{4}} \frac{\sec^2 \phi - 2 \tan \phi}{1 + \cos 4 \phi}.$		$\frac{1}{2}.$
6. $\lim_{x \to \frac{\pi}{2}} \frac{\log \sin x}{(\pi - 2x)^2}.$		$-\frac{1}{8}.$	14. $\lim_{z \to a} \frac{az - z^2}{a^4 - 2a^3z + 2 az^3 - z^4}.$		$+\infty.$
7. $\lim_{x \to 0} \frac{a^x - b^x}{x}.$		$\log \frac{a}{b}.$	15. $\lim_{x \to 2} \frac{(e^x - e^2)^2}{(x - 4)e^x + e^2 x}.$		6e4.
8. $\lim_{r \to a} \frac{r^3 - ar^2 - a^2 r + a^3}{r^2 - a^2}.$		0.

16. $\lim_{x \to 1} \frac{x^2 + x - 2}{x^2 - 1}.$	18. $\lim_{x \to 0} \frac{\sin 2x}{x}.$	20. $\lim_{x \to 1} \frac{\log \cos (x - 1)}{1 - \sin \frac{\pi x}{2}}.$
17. $\lim_{x \to -2} \frac{x^3 + 8}{x^5 + 32}.$	19. $\lim_{x \to 0} \frac{x - \sin x}{x^3}.$	21. $\lim_{x \to 0} \frac{\tan x - \sin x}{\sin^3 x}.$

112. Evaluation of the indeterminate form $\tfrac{\infty}{\infty}$. In order to find when

	$\lim_{x \to a} \frac{f(x)}{F(x)}$
when	$\lim_{x \to a} f(x) = \inf \text{and} \lim_{x \to a} F(x) = \infty,$
that is, when for x = a the function
	$\frac{f(x)}{F(x)}$
assumes the indeterminate form
	$\frac{\infty}{\infty}$

we follow the same rule as that given in §111 for evaluating the indeterminate form $\tfrac{0}{0}$. Hence

Rule for evaluating the indeterminate form $\tfrac{\infty}{\infty}$. Differentiate the numerator for a new numerator and the denominator for a new denominator. The value of this new fraction for the assigned value of the variable will be the limiting value of the original fraction.

A rigorous proof of this rule is beyond the scope of this book and is left for more advanced treatises.

Illustrative Example 1. Evaluate $\tfrac{\log x}{\csc x}$ for x = 0.

Solution.	$\frac{f(0)}{F(0)} = \left. \frac{\log(x)}{\csc(x)} \right]_{x = 0} = \frac{-\infty}{\infty}.$ ∴ indeterminate.
	$\frac{f'(0)}{F'(0)} = \left. \frac{\frac{1}{x}}{-\csc x \cot x} \right]_{x = 0} = \left. -\frac{\sin^2 x}{x \cos x} \right]_{x = 0} = \frac{0}{0}.$ ∴ indeterminate.
	$\frac{f''(0)}{F''(0)} = \left. -\frac{2 \sin x \cos x}{\cos x - x \sin x} \right]_{x = 0} = -\frac{0}{1} = 0.$ Ans.

113. Evaluation of the indeterminate form $0 \cdot \infty$. If a function $f(x) \cdot \phi(x)$ takes on the indeterminate form $0 \cdot \infty$ for x = a, we write the given function

$f(x) \cdot \phi(x) = \frac{f(x)}{ \frac{1}{\phi(x)} }$ $\left( \text{or} = \frac{\phi(x)}{\frac{1}{f(x)}} \right)$

so as to cause it to take on one of the forms $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$, thus bringing it under §111 or §112. Illustrative Example 1. Evaluate sec 3xcos 5x for $x = \tfrac{\pi}{2}$.

Solution. $\left. \sec 3x \cos 5x \right]_{x = \frac{\pi}{2}} = \infty \cdot 0.$ ∴ indeterminate.

Substituting $\tfrac{1}{\cos 3x}$ for sec 3x, the function becomes $\tfrac{\cos 5x}{\cos 3x} = \tfrac{f(x)}{F(x)}$.

$\frac{ f(\frac{\pi}{2}) }{ F(\frac{\pi}{2}) }$	$= \left. \frac{\cos 5x}{\cos 3x} \right]_{x = \frac{\pi}{s}} = \frac{0}{0}.$ indeterminate.
$\frac{ f'(\frac{\pi}{2}) }{ F'(\frac{\pi}{2}) }$	$= \left. \frac{-\cos x}{- \sin x} \right]_{x = \frac{\pi}{s}} = \frac{0}{-1} = 0.$ Ans.

114. Evaluation of the indeterminate form $\infty -\infty$. It is possible in general to transform the expression into a fraction which will assume either the form $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$.

Illustrative Example 1. Evaluate sec x − tan x for $x = \tfrac{\pi}{2}$.

Solution. $\left. \sec x - \tan x \right]_{x= \tfrac{\pi}{2}} = \infty - \infty.$ ∴ indeterminate.

By Trigonometry, $\sec x - \tan x = \tfrac{1}{\cos x} - \tfrac{\sin x}{\cos x} = \tfrac{1 - \sin x}{\cos x} = \tfrac{f(x)}{F(x)}$.

$\frac{ f(\frac{\pi}{2}) }{ F(\frac{\pi}{2}) }$	$= \left. \frac{1 - \sin x}{\cos x} \right]_{x = \frac{\pi}{2}} = \frac{1 - 1}{0} = \frac{0}{0}.$ ∴ indeterminate.
$\frac{ f'(\frac{\pi}{2}) }{ F'(\frac{\pi}{2}) }$	$= \left. \frac{-\cos x}{-\sin x} \right]_{x = \frac{\pi}{2}} = \frac{0}{-1} = 0.$ Ans.

EXAMPLES

Evaluate the following expressions by differentiation:

1. $\lim_{x \to \infty} \frac{ax^2 + b}{cx^2 + d}.$	Ans.	$\frac{a}{c}.$	6. $\lim_{x \to 0} \frac{\log \sin 2x}{\log \sin x}.$	Ans.	1.
2. $\lim_{x \to 0} \frac{\cot x}{\log x}.$		$-\infty.$	7. $\lim_{\theta \to \frac{\pi}{2}} \frac{\tan \theta}{\tan 3\theta}.$		3.
3. $\lim_{x = \infty} \frac{\log x}{x^n}.$		0.	8. $\lim_{\phi \to \frac{\pi}{2}} \frac{\log \left( \phi - \frac{\pi}{2} \right)}{\tan \phi}.$		0.
4. $\lim_{x \to \infty} \frac{x^2}{e^x}.$		0.	9. $\lim_{x \to 0} \frac{\log x}{\cot x}.$		0.
5. $\lim_{x \to \infty} \frac{e^x}{\log x}.$		$\infty.$	10. $\lim_{x \to 0} x \log \sin x.$		0.

11. $\lim_{x \to 0} x \cot \pi x.$	Ans.	$\frac{1}{\pi}.$	18. $\lim_{x \to 1} \left[ \frac{2}{x^2 - 1} - \frac{1}{x - 1} \right].$	Ans.	$-1\frac{1}{2}$
12. $\lim_{y \to \infty} \frac{y}{e^{ay}}.$		0.	19. $\lim_{x \to 1} \left[ \frac{1}{\log x} - \frac{x}{\log x} \right].$
13. $\lim_{x \to \frac{\pi}{2}} (\pi - 2x) \tan x.$		2.	20. $\lim_{\theta \to \frac{\pi}{2}} [\sec \theta - \tan \theta].$		0.
14. $\lim_{x \to \infty} x \sin \frac{a}{x}.$		a.	21. $\lim_{\phi \to 0} \left[ \frac{2}{\sin^2 \phi} - \frac{1}{1 - \cos \phi} \right].$		$\frac{1}{2}.$
15. $\lim_{x \to 0} x^n \log x.$ [n positive.]		0.	22. $\lim_{y \to 1} \left[ \frac{y}{y - 1} - \frac{1}{\log y} \right].$		$\frac{1}{2}.$
16. $\lim_{\theta \to \frac{\pi}{4}} (1 - \tan \theta) \sec 2\theta.$		1.	23. $\lim_{z \to 0} \left[ \frac{\pi}{4z} - \frac{\pi}{2z(e^{\pi z} + 1)} \right].$		$\frac{\pi^2}{8}.$
17. $\lim_{\phi \to a} (a^2 - \phi^2) \tan \frac{\pi \phi}{2a}.$		$\frac{4 a^2}{\pi}.$

115. Evaluation of the indeterminate forms $0^0, 1^{\infty}, \infty^0$. Given a function of the form

	f(x)ϕ(x).
In order that the function shall take on one of the above three forms, we must have for a certain value of x
	f(x) = 0,ϕ(x) = 0,giving00;
or,	$f(x) = 1, \phi(x) = \infty, \text{giving} 1^{\infty}$;
or,	$f(x) = \infty, \phi(x) = 0, \text{giving} {\infty}^0$,
Let	y = f(x)ϕ(x);
taking the logarithm of both sides,
	log y = ϕ(x)log f(x).
In any of the above cases the logarithm of y (the function) will take on the indeterminate form
	$0 \cdot \infty.$

Evaluating this by the process illustrated in §113 gives the limit of the logarithm of the function. This being equal to the logarithm of the limit of the function, the limit of the function is known.

Illustrative Example 1. Evaluate x^x when x = 0.

Solution. This function assumes the indeterminate form 0⁰ for x = 0.

Let	$\ y$	$=\ x^x;$
then	$\ \log y$	$= x \log x = 0 \cdot -\infty,$	when x = 0.
By § 113,	$\ \log y$	$\frac{\log x}{\frac{1}{x}} = \frac{-\infty}{\infty},$	when x = 0.

By § 112,

$\ \log y$

$\frac{\frac{1}{x}}{-\frac{1}{x^2}} = -x = 0,$

when x = 0.

Since y = x^x, this gives log_ex^x = 0; i.e., x^x = 1. Ans.

Illustrative Example 2. Evaluate $(1 + x)^{\tfrac{1}{x}}$; when x = 0.

Solution. This function assumes the indeterminate form $1^{\infty}$ for x = 0.

Let	$\ y$	$= (1 + x)^{\frac{1}{x}};$
then	$\ \log y$	$= \frac{1}{x} \log(1 + x) = \infty \cdot . 0,$	when x = 0.
By § 113,	$\ y$	$= \frac{\log(1 + x)}{x} = \frac{0}{0},$	when x = 0.
By § 111,	$\ y$	$= \frac{\frac{1}{1 + x}}{1} = \frac{1}{1 + x} = 1,$	when x = 0.

since $y = (1 + x)^{\frac{1}{1 + x}}$, this gives $log_e (1 + x)^{\frac{1}{x}} = 1$; i.e. $(1 + x)^{\frac{1}{x}} = e$. Ans.

Illustrative Example 3. Evaluate (cot x)^sin x for x = 0.

Solution. This function assumes the indeterminate form $\infty^0$ for x = 0.

Let	$\ y$	$=\ (\cot x)^{\sin x};$
then	$\ \log y$	$=\ \sin x \log \cot x = 0 \cdot \infty,$	when x = 0.
By § 113,	$\ \log y$	$= \frac{\log \cot x}{\csc x} = \frac{\infty}{\infty},$	when x = 0.
By § 112,	$\ \log y$	$= \frac{\frac{-\csc^2 x}{\cot x}}{-\csc x \cot x} = \frac{\sin x}{\cos^2 x} = 0,$	when x = 0.

EXAMPLES

Evaluate the following expressions by differentiation:

1. $\lim_{x \to 1} x^{\frac{1}{1 - x}}.$	Ans.	$\frac{1}{e}.$	7. $\lim_{x \to 0} (e^x + x)^{\frac{1}{x}}.$	Ans.	e2.
2. $\lim_{x \to 0} \left( \frac{1}{x} \right)^{\tan x}.$		1.	8. $\lim_{x \to 0} (\cot x)^{\frac{1}{\log x}}.$		$\frac{1}{e}.$
3. $\lim_{\theta \to \frac{\pi}{2}} (\sin \theta)^{\tan \theta}.$		1.	9. $\lim_{z \to 0} (1 + nz)^{\frac{1}{z}}.$		en.
4. $\lim_{y \to \infty} \left( 1 + \frac{a}{y} \right)^y.$		ea.	10. $\lim_{\phi \to 1} \left( \tan \frac{\pi \phi}{4} \right)^{\tan \frac{\pi \phi}{2}}.$		$\frac{1}{e}.$
5. $\lim_{x \to 0} (1 + \sin x)^{\cot x}.$		e.	11. $\lim_{\theta \to 0} (\cos m\theta)^{\frac{n}{\theta^2}}.$		$e^{-\frac{1}{2} nm^2}.$
6. $\lim_{x \to \infty} \left( \frac{2}{x} + 1 \right)^x.$		e2.	12. $\lim_{x \to 0} (\cot x)^x.$		1.
			13. $\lim_{x \to a} \left( 2 -\frac{x}{a} \right)^{\tan \frac{\pi x}{2a}}.$		$e^{\frac{2}{\pi}}.$

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