﻿ ﻿EVALUATION OF INDETERMINATE FORMS - Indeterminate Forms - The Calculus Primer

## The Calculus Primer (2011)

### Chapter 34. EVALUATION OF INDETERMINATE FORMS

9—4.Indeterminate Forms. It often happens that for a particular value of the independent variable, a function may take on a corresponding value which is indeterminate, such as , 0· ∞, ∞ – ∞, 00, l, or ∞0.

For example, what is the value of y in

when x = − 3? Direct substitution gives

which is undefined. However, if we first write

then y = −3 − 3 = −6 when x = −3.

Or, consider another example: find the value of

Direct substitution gives the value , which again is indeterminate, or undefined. By an algebraic transformation, however, we may write

Such algebraic transformations, however, do not constitute general methods for evaluating indeterminate forms. For this purpose we can use the process of differentiation, as shown in what follows.

9—5.The Indeterminate Form . Let us consider a function of the form such that f(a) = 0 and F (a) = 0; in other words, suppose that the function takes on the indeterminate form when a is substituted for x. The problem, then, is to determine the value of

The curves y = f(x) and y = F(x) must intersect at (a,0), since f(a) = 0 and F (a) =0 by hypothesis. Let us now employ the law of the mean:

f(x) = f(a) + (xa)f′(x1),wherea < x1 < x;

and F(x) = F(a) + (xa)F′(x2), wherea < x2 < x.

Dividing these two equations, we have:

the terms f(a) and F(a) vanishing, since each is equal to zero. Finally, let xa; then x1a and x2a; hence

This last relation, then, suggests a method of procedure to be used in evaluating a function which leads to the form .

To evaluate the indeterminate form , differentiate the numerator to obtain a new numerator, and the denominator to obtain a new denominator. The value of this new fraction for the assigned value of the variable will be the desired limiting value of the original fraction.

NOTE 1. If, after differentiation, it should turn out that both f′(a) and F′(a) are equal to zero, the rule may be applied again to this ratio; thus

In like manner, if f″(a) = 0 and F″(a) = 0, then apply the rule repeatedly, until the indeterminate form does not appear; thus

NOTE 2. It can be shown that the rule is also valid when a = ∞.

NOTE 3. When applying the rule, the expression should not be differentiated as a fraction; each term of the fraction should be differentiated separately.

EXAMPLE 1.Evaluate

Solution. Let f(x) = x2 − 9, and F(x) = 2x2 + x − 21.

Solution. Let f(x) = 1 – cos x, and F(x) = sin x.

Solution. Let f(x) = ex − 2 sin xex, and F (x) = x − sin x.

Substituting when x = 0:

EXERCISE 9—1

Evaluate each of the following by differentiating:

9—6.The Indeterminate Form . To determine the value of an expression whose limit reduces to , we proceed according to the same rule as used when evaluating the form . A rigorous proof of the validity of the rule for the case must be left, however, for more advanced study.

EXAMPLE 1.Evaluate when x = 0.

Solution. Let f(x) = log x, and F(x) = .

EXAMPLE 2.Find the limit of when x = ∞.

Solution. Let f(x) = x2, and F(x) = ex.

9—7.Evaluation of the Form 0·∞. When a function such as f(x)·ø(x) takes the indeterminate form 0· ∞ when x = a, it may be evaluated by writing

In this way it will take either the form or , which may then be evaluated by the method of §9—5 or §9—6.

EXAMPLE 1.Evaluate x2ex3 when x = ∞.

Solution. Let f(x) = x2, and ø(x) = e−x3.

When x = ∞, f(x)·ø(x) = ∞·0; indeterminate.

Writing f(xø(x) as the function becomes

EXAMPLE 2.Evaluate sec x cos 3x when

9—8.Evaluation of the Form ∞ − ∞. An expression which reduces to this form can usually be transformed into a fraction that will assume the form or .

EXAMPLE 1.Find the value of

Write the given expression as

Write the given expression as

EXERCISE 9—2

Evaluate each of the following:

9—9.Evaluation of the Forms 00, 1, and ∞0. A function of the form f(x)ø(x) may yield indeterminate forms as follows:

(A)if f(x) = 0, and ø(x) = 0, we obtain 00;

(B)if f(x) = 1, and ø(x) = ∞, we obtain l;

(C)if f(x) = ∞, and ø(x) = 0, we obtain ∞0.

In such cases, the indeterminate forms are evaluated by a logarithmic transformation. Thus, let

y = f(x)ø(x);

thenlog y = ø(x) log f(x).

Now we have the indeterminate form 0· ∞; for

in (A), log y = 0· (log 0) = 0· (− ∞ ).

in (B), log y = ∞ · (log 1) = ∞ · 0.

in (C), log y = 0 · (log ∞) = 0 · ∞.

The expression ø(x) log f(x) may therefore be evaluated as in §9—7; the limit so found, however, is the limit of the logarithm of the desired function. But the limit of the logarithm of a function equals the logarithm of the limit of the function. Thus, if we know lim loge y = a, then y = ea.

EXAMPLE 1.Find the value of

(1 + kx)1/x.

Solution. (1 + kx)1/x = 1; indeterminate.

Puty = (1 + kx)1/x;

thenlog y = log (1 + kx), and

Since lim loge y = k, then y = ek; that is, (1 + kx)1/x = ek. In other words, since y = (1 + kx)1/x, this gives loge (1 + kx)1/x = k; that is, (1 + kx)1/x = ek.

EXAMPLE 2.Evaluate when x = ∞.

EXAMPLE 3.Evaluate (cot x)x.

Solution. (cot x)x = ∞0; indeterminate.

Put y = (cot x)x;

thenlog y = x log cot x.

(x log cot x) = 0· ∞.

To find (x log cot x), write the expression as

Hence, since lim loge y = 0, then y = e0 = 1; that is, y = (cot x)x = 1. In other words, since y = (cot x)x, this gives loge (cot x)x = 1; that is, (cot x)x = 1.

9—10.The Form 0. It is interesting to note that the form 0 is not indeterminate; it is always equal to zero. This can be explained as follows. In the function y = f(x)F(x), let f(x) = 0 and F(x) = ∞. Then,

log y = F(x) log f(x),

orlog y = (∞)·(− ∞ ) = − ∞;

since lim loge y = − ∞ y = e−∞ = = 0. Hence 0 = 0.

EXERCISE 9—3

Evaluate each of the following:

EXERCISE 9—4

Review

1. Find the equation of the tangent to the curve y2 + 3x − 2y + 4 = 0 at the point (2,2).

2. Find the equation of the tangent line to the parabola y2 = 8x + 12, parallel to the line 2x − 2y = 3.

3. Find the value of:

(a) (cot x)1/logx.

(b)lim (x − l)a/log sin πx.

(c) (1 − ez)x.

4. Find and for the equation x2 + xy + y2 = 0.

5. A point moves so that its distance S at any instant t is given by S = t5 − 5t4. At what time t is the acceleration changing most rapidly?

6. Sand is falling onto a conical pile at the rate of 18 cu. ft. per minute. The radius of the pile is always one-third of its altitude. How fast is the altitude of the pile increasing when the pile is 6 feet high?

7. Find the point of inflection of the curve y = x3 + 3x2 + 12.

8. What is the curvature at any point of the curve y = e2x?

9. Find the radius of curvature of:

(a) at any point.

(b) y = x3, at the point where x = 1.

10. Prove that the sum of any positive real number and its reciprocal is equal to or greater than 2.

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