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ADDITIONAL NOTES AND EXERCISES FOR 4.
MIKE WILLS
- Indeterminate Forms
Definition 1.1. The set of the extended real numbers is
(1.1) R ∪ {∞, −∞}
where ∞ and −∞ are symbols which satisfy the inequality
(1.2) −∞ < a < ∞
for every real number a.
An expression like 20 is undefined in the real numbers. However, in the extended real numbers the expression 20 has two reasonable interpretations: either ∞ or −∞. In other words, in the extended real numbers 20 is determined up to sign. There are still issues of course, but the point is that if f (x) → 0 as x → 0 then
if lim x→a
f (x)
exists it must be either ∞ or −∞. By contrast, an indeterminant
form is an expression for which a whole range of values can be assigned with equal (lack of) validity. For example, 00 does not make sense even in the context of the extended real numbers; it matters how we approach zero in the numerator and the denominator. Indeterminant forms arise most naturally in the evaluation of limits: For example, the observation
(1.3) a = lim x→ 0
ax x
lim x→ 0 ax lim x→ 0 x
illustrates that 00 can in some sense take every possible real value. When evaluating a limit where an indeterminant form is present we cannot blindly misapply limit laws and hope to get a reasonable result. This is where l’Hˆopital’s rule comes into play.
- L’Hˆopital’s rule and the indeterminate forms 00 and ∞∞ L’Hˆopital’s rule gives us an excellent way of computing limits of ratios of func- tions whose numerator and denominator both converge/diverge either to zero or to ∞. The text discusses how one can use l’Hˆopital’s rule. The point of this section is to highlight the pitfalls that occur when the rule is blindly misapplied. Here are some things to bear in mind. First, in many cases it is more effective and efficient to use algebra as was done in the example in the last section. Second, l’Hˆopital’s rule is ineffective when we are not working with an indeter- minate form: 1
2 MIKE WILLS
For example,
(2.1) 0 = lim x→ 0
sin x 1 + cos x
= − lim x→ 0
cos x sin x
The right hand side is undefined in the extended real numbers. a Hence, in general, the limit of the ratio of the original functions is not necessarily the limit of the ratio of the derivatives even when both limits exist. Third, and more subtly, l’Hˆopital’s rule may be inapplicable for technical reasons. Consider the following expression:
(2.2) lim x→ 0
sin x x
Here, the ratio of the limits is the indeterminate form 00 which means l’Hˆopital’s rule is potentially applicable. Indeed,
(2.3) lim x→ 0
sin x x
= lim x→ 0
cos x 1
The problem is that to prove that the derivative of sin x is cos x we had to show
that lim x→ 0
sin x x
= 1. Thus, using l’Hˆopital’s rule to evaluate our limit is begging the
question. Fourth, the limit may exist and we may have an appropriate indeterminate form but l’Hˆopital’s rule is inapplicable either because the functions under study are not differentiable or because the limit of the ratio of the derivatives does not exist.
Example 2.1. Let
(2.4) g(x) =
0 x /∈ Q x x ∈ Q
Then
(2.5) lim x→ 0
g(x) = lim x→ 0
x = 0
but l’Hˆopital’s rule is inapplicable because the denominator is not differentiable. Nevertheless,
(2.6) lim x→ 0
x^2 g(x)
Fifth, l’Hˆopital’s rule may be applicable but not yield any useful information. As a somewhat silly example, try applying l’Hˆopital’s rule to
lim x→∞
ex ex^
A somewhat more substantive example is given in the exercises.
- The indeterminate form ∞ − ∞ Although l’Hˆopital’s rule is not directly applicable to indeterminant forms of this type sometimes one can still use the rule by converting the difference into a quotient. The idea is best illustrated with an example:
4 MIKE WILLS
Since ln is continuous, ln(f ) → −∞ as x → 0. Since exp is continuous,
(5.2) lim x→a eg(x) ln^ f^ (x)^ = e
lim x→a g(x) ln f (x)
which gives the indeterminant form 0 · ∞. Similarly, if f (x) > 0, f → ∞, and g → 0 as x → a then ln f → ∞ as x → a and so we again have the indeterminant form 0 · ∞. If f → 1 and g → ±∞ as x → a then ln f → 0 as x → a and the same calculation gives the indeterminant form ∞ · 0. One example should illustrate the method.
Example 5.1. We compute lim x↓ 0 xx
4 :
(5.3) lim x↓ 0
xx
4 = lim x↓ 0
ex
(^4) ln x = e
lim x↓ 0
ln x x−^4 = e
−lim x↓ 0
1 x 4 x−^5 = e
−lim x↓ 0
x^4 (^4) = 1.
- Exercises In exercises one through ten, evaluate the limits if they exist. If they do not exist, explain why.
Exercise 6.1.
lim x→ 0
sinh x sin x Exercise 6.2. lim x→∞ (x (^13) − x^3 )
Exercise 6.3.
lim x→π
sin x cos x + 1
Exercise 6.4. lim x→∞ xnex^ (n ∈ N)
Exercise 6.5. lim x→∞ x
(^1) x
Exercise 6.6.
lim x→ 1
x^2 − 1 x − 1
Exercise 6.7.
lim x↑ π 2
tan x x − π 2
Exercise 6.8.
lim x→−∞
ln(ln |x|) x
Exercise 6.9.
lim x→ 0
∫ (^) x 0 cos(t
(^2) )dt x^3
Exercise 6.10.
lim x→∞
x ln x 3 x^2 − 1
ADDITIONAL NOTES AND EXERCISES FOR 4.8 5
Exercise 6.11. Let f (x) = xx−+sinsin^ xx. Using trigonometric identities, compute
(6.1) lim x→ 0 f (x).
Explain why l’Hˆopital’s rule is inapplicable, even though we have the indeterminant form ∞∞ and the functions are differentiable.
Exercise 6.12. Let f (x) =
√ x^2 + x. Compute (6.2) lim x→∞ f (x).
Explain why l’Hˆopital’s rule is ineffective.
The next two exercises show that the undefined expression 0∞^ is not an indeter- minant form.
Exercise 6.13. Suppose that f (x) > 0 and that f (x) → 0, g(x) → ∞ as x → a. Compute
(6.3) lim x→a f (x)g(x).
Exercise 6.14. Suppose that f (x) > 0 and that f (x) → 0, g(x) → −∞ as x → a. Compute
(6.4) lim x→a f (x)g(x).
