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31.1 Limit of indeterminate type
Some limits for which the substitution rule does not apply can be found by using inspection. For instance,
lim x→ 0
cos x x^2
about 1 small pos.
On the other hand, we have seen (8) that inspection cannot be used to find the limit of a fraction when both numerator and denominator go to 0. The examples given were
lim x→ 0 +
x^2 x
, lim x→ 0 +
x x^2
, lim x→ 0 +
x x
In each case, both numerator and denominator go to 0. If we had a way to use inspection to decide the limit in this case, then it would have to give the same answer in all three cases. Yet, the first limit is 0, the second is ∞ and the third is 1 (as can be seen by canceling x’s).
We say that each of the above limits is indeterminate of type 00. A useful way to remember that one cannot use inspection in this case is to imagine that the numerator going to 0 is trying to make the fraction small, while the denominator going to 0 is trying to make the fraction large. There is a struggle going on. In the first case above, the numerator wins (limit is 0); in the second case, the denominator wins (limit is ∞); in the third case, there is a compromise (limit is 1).
Changing the limits above so that x goes to infinity instead gives a different indeterminate type. In each of the limits
xlim→∞
x^2 x , (^) xlim→∞
x x^2 , (^) xlim→∞
x x
both numerator and denominator go to infinity. The numerator going to infinity is trying to make the fraction large, while the denominator going to infinity is trying to make the fraction small. Again, there is a struggle. Once again, we can cancel x’s to see that the first limit is ∞ (numerator wins), the second limit is 0 (denominator wins), and the third limit is 1 (compromise). The different answers show that one cannot use inspection in this case. Each of these limits is indeterminate of type ∞∞.
Sometimes limits of indeterminate types 00 or ∞∞ can be determined by using some algebraic technique, like canceling between numerator and denominator
as we did above (see also 12). Usually, though, no such algebraic technique suggests itself, as is the case for the limit
lim x→ 0
x^2 sin x
which is indeterminate of type 00. Fortunately, there is a general rule that can be applied, namely, l’Hˆopital’s rule.
31.2 L’Hˆopital’s rule
L’Hˆopital’s rule. If the limit
lim
f (x) g(x)
is of indeterminate type 00 or ±∞±∞ , then
lim
f (x) g(x) = lim
f ′(x) g′(x)
provided this last limit exists. Here, lim stands for lim x→a , lim x→a±
or lim x→±∞
The pronunciation is l¯o-p¯e-t¨al. Evidently, this result is actually due to the mathematician Bernoulli rather than to l’Hˆopital. The verification of l’Hˆopital’s rule (omitted) depends on the mean value theorem.
31.2.1 Example Find lim x→ 0
x^2 sin x
Solution As observed above, this limit is of indeterminate type 00 , so l’Hˆopital’s rule applies. We have
lim x→ 0
x^2 sin x
l’H = lim x→ 0
2 x cos x
where we have first used l’Hˆopital’s rule and then the substitution rule.
The solution of the previous example shows the notation we use to indicate the type of an indeterminate limit and the subsequent use of l’Hˆopital’s rule.
31.2.2 Example Find lim x→−∞
3 x − 2 ex^2
Sometimes repeated use of l’Hˆopital’s rule is called for:
31.4.2 Example Find (^) xlim→∞
3 x^2 + x + 4 5 x^2 + 8x
Solution We have
lim x→∞
3 x^2 + x + 4 5 x^2 + 8x
) (^) l’H = lim x→∞
6 x + 1 10 x + 8
) (^) l’H = lim x→∞
For the limit at infinity of a rational function (i.e., polynomial over polynomial) as in the preceding example, we also have the method of dividing numerator and denominator by the highest power of the variable in the denominator (see 12). That method is probably preferable to using l’Hˆopital’s rule repeatedly, especially if the degrees of the polynomials are large. Sometimes though, we have no alternate approach:
31.4.3 Example Find lim x→ 0
ex^ − 1 − x − x^2 / 2 x^3
Solution We have
lim x→ 0
ex^ − 1 − x − x^2 / 2 x^3
l’H = lim x→ 0
ex^ − 1 − x 3 x^2
l’H = lim x→ 0
ex^ − 1 6 x
l’H = lim x→ 0
ex 6
e^0 6
There are other indeterminate types, to which we now turn. The strategy for each is to transform the limit into either type 00 or ±∞±∞ and then use l’Hˆopital’s rule.
31.5 Indeterminate product
Type ∞ · 0. The limit lim x→∞ xe−x^ (∞ · 0)
cannot be determined by using inspection. The first factor going to ∞ is trying to make the expression large, while the second factor going to 0 is trying to make the expression small. There is a struggle going on. We say that this limit is indeterminate of type ∞ · 0.
The strategy for handling this type is to rewrite the product as a quotient and then use l’Hˆopital’s rule.
31.5.1 Example Find lim x→∞ xe−x.
Solution As noted above, this limit is indeterminate of type ∞ · 0. We rewrite the expression as a fraction and then use l’Hˆopital’s rule:
xlim→∞ xe−x^ =^ xlim→∞
x ex
l’H = lim x→∞
ex
large pos.
31.5.2 Example Find lim x→ 0 +
(cot 2x)(sin 6x).
Solution First
lim x→ 0 +^
cot 2x = lim x→ 0 +
cos 2x sin 2x
about 1 small pos.
Therefore, the given limit is indeterminate of type ∞·0. We rewrite as a fraction and then use l’Hˆopital’s rule:
lim x→ 0 +
(cot 2x)(sin 6x) = lim x→ 0 +
sin 6x tan 2x
l’H = lim x→ 0 +
6 cos 6x 2 sec^2 2 x =
31.6 Indeterminate difference
Type ∞ − ∞. The substitution rule cannot be used on the limit
lim x→ π 2 −
(tan x − sec x) (∞ − ∞)
since tan π/2 is undefined. Nor can one determine this limit by using inspec- tion. The first term going to infinity is trying to make the expression large and positive, while the second term going to negative infinity is trying to make the
cannot be determined by using inspection. The base going to infinity is trying to make the expression large, while the exponent going to 0 is trying to make the expression equal to 1. There is a struggle going on. We say that this limit is indeterminate of type ∞^0.
The strategy for handling this type (as well as the types 1∞^ and 0^0 yet to be introduced) is to first find the limit of the natural logarithm of the expression (ultimately using l’Hˆopital’s rule) and then use an inverse property of logarithms to get the original limit.
31.7.1 Example Find lim x→∞ x^1 /x.
Solution As was noted above, this limit is of indeterminate type ∞^0. We first find the limit of the natural logarithm of the given expression:
lim x→∞ ln x^1 /x^ = lim x→∞ (1/x) ln x (0 · ∞)
= lim x→∞
ln x x
l’H = lim x→∞
1 /x 1 = 0.
Therefore, lim x→∞ x^1 /x^ = lim x→∞ eln^ x
1 /x = e^0 = 1,
where we have used the inverse property of logarithms y = eln^ y^ and then the previous computation. (The next to the last equality also uses continuity of the exponential function.)
Type 1 ∞. The limit
lim x→∞
x
)x (1∞)
cannot be determined by using inspection. The base going to 1 is trying to make the expression equal to 1, while the exponent going to infinity is trying to make the expression go to ∞ (raising a number greater than 1 to ever higher powers produces ever larger results). We say that this limit is indeterminate of type 1∞.
31.7.2 Example Find lim x→∞
x
)x .
Solution As was noted above, this limit is indeterminate of type 1∞. We first
find the limit of the natural logarithm of the given expression:
lim x→∞ ln
x
)x = lim x→∞ x ln
x
= (^) xlim→∞
ln (1 + 1/x) x−^1
l’H = lim x→∞
1 + 1/x
−x−^2
−x−^2 = lim x→∞
1 + 1/x
Therefore,
lim x→∞
x
)x = lim x→∞ eln(1+^
(^1) x )x = e^1 = e.
Type 00. The limit lim x→ 0 (tan x)x^
cannot be determined by using inspection. The base going to 0 is trying to make the expression small, while the exponent going to 0 is trying to make the expression equal to 1. We say that this limit is indeterminate of type 0^0.
31.7.3 Example Find lim x→ 0 +
(tan x)x.
Solution As was noted above, this limit is indeterminate of type 0^0. We first find the limit of the natural logarithm of the given expression:
lim x→ 0 +^
ln(tan x)x^ = lim x→ 0 +^
x ln tan x
= lim x→ 0 +
ln tan x x−^1
l’H = lim x→ 0 +
cot x sec^2 x −x−^2
= lim x→ 0 +
−x^2 sin x cos x
l’H = lim x→ 0 +
− 2 x cos^2 x − sin^2 x =
Therefore, lim x→ 0 +
(tan x)x^ = lim x→ 0 +^
eln(tan^ x)
x = e^0 = 1.
31 – 1 Find lim x→ 0
e^3 x^ − 1 x
31 – 2 Find lim x→∞
ln x x^2
31 – 3 Find lim x→ 0 +^
x ln x.
31 – 4 Find lim x→ 0
ex^ − x − 1 cos x − 1
31 – 5 Find lim x→ 0 +
sin x
x
31 – 6 Find lim x→π/ 2
1 − sin x csc x
31 – 7 Find lim x→ 0 +
(sin x)x.
31 – 8 Find lim x→∞ (ex^ + x)^1 /x.
