## The Calculus Primer (2011)

### Part IX. Indeterminate Forms

### 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· ∞, ∞ – ∞, 0^{0}, 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*) + (*x* − *a*)*f′*(*x*_{1})*,*where*a < x*_{1} *< x*;

and *F*(*x*) *= F*(*a*) *+* (*x* − *a*)*F′*(*x*_{2}), where*a* < *x*_{2} < *x*.

Dividing these two equations, we have:

the terms *f*(*a*) and *F*(*a*) vanishing, since each is equal to zero. Finally, let *x* → *a*; then *x*_{1} → *a* and *x*_{2} → *a*; 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*) = *x*^{2} − 9, and *F*(*x*) = 2*x*^{2} *+ x* − 21.

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

*Solution.* Let *f*(*x*) = *e ^{x}* − 2 sin

*x*−

*e*

^{′x}, 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*) = *x*^{2}, and *F*(*x*) = *e ^{x}.*

**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 *x*^{2}*e*^{−x3} when *x* = ∞.

*Solution. Let f*(*x*) = *x*^{2}, and *ø*(*x*) = *e ^{−x}*

^{3}.

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

Writing *f*(*x*)·*ø*(*x*) as the function becomes

EXAMPLE 2.Evaluate sec *x* cos 3*x* 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 0 ^{0}, 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 0^{0};

(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 log_{e}*y* = *a*, then *y = e ^{a}.*

EXAMPLE 1.Find the value of

(1 + *kx*)^{1/x}.

*Solution.* (1 + *kx*)^{1/x} = 1^{∞}; indeterminate.

Put*y* = (1 + *kx*)^{1/x};

thenlog *y = * log (1 + *kx*), and

Since lim log_{e}*y* = *k,* then *y* = *e ^{k}*; that is, (1 +

*kx*)

^{1/x}=

*e*. In other words, since

^{k}*y*= (1 +

*kx*)

^{1/x}, this gives log

*(1 +*

_{e}*kx)*

^{1/x}=

*k*; that is, (1 +

*kx*)

^{1/x}=

*e*.

^{k}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 log_{e}*y* = 0, then *y* = *e*^{0} = 1; that is, *y* = (cot *x*)* ^{x}* = 1. In other words, since

*y*= (cot

*x*)

*, this gives log*

^{x}*(cot*

_{e}*x*)

*= 1; that is, (cot*

^{x}*x*)

*= 1.*

^{x}**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 log_{e}*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 *y*^{2} + 3*x* − 2*y* + 4 = 0 at the point (2,2).

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

**3.** Find the value of:

(a) (cot *x)*^{1/logx}.

(b)lim (*x* − l)^{a}^{/log sin πx}.

(c) (1 − *e ^{z}*)

^{x}.**4.** Find and for the equation *x*^{2} + *xy* + *y*^{2} = 0.

**5.** A point moves so that its distance *S* at any instant *t* is given by *S = t*^{5} − 5*t*^{4}. 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* = *x*^{3} *+* 3*x*^{2} + 12.

**8.** What is the curvature at any point of the curve *y* = *e ^{2x}?*

**9.** Find the radius of curvature of:

(a) at any point.

(b) *y = x*^{3}, 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.