3.6 Indeterminate Forms and L’Hospital’s Rule
3.6
Brian E. Veitch
Indeterminate Forms and L’Hospital’s Rule
From calculus I, we used a geometric approach to show
sin(x)
=1
x→0
x
lim
We needed to use a geometric approach back then because we didn’t know what to do
0
with and we couldn’t use one of our limit methods (like using conjguates, factoring, etc.).
0
In general, if you have a limit of the form
f (x)
x→a g(x)
lim
and f (x) → 0 and g(x) → 0 as x → a, then the limit may or may not exist. That’s
for us to find out.
Example 3.34. Indeterminate Form of Type
0
0
sin(x)
x→0
x
lim
because limx→0 sin(x) = 0 and limx→0 x = 0
Example 3.35. Indeterminate Form of Type
lim
x→∞
∞
∞
ln x − 1
x−1
104
3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
Just to be clear, this isn’t an indeterminate form because x → ∞. It’s because
lim ln x − 1 = ∞ and lim x − 1 = ∞
x→∞
3.6.1
x→∞
L’Hosptial’s Rule
Suppose f and g are differentiable and g 0 (x) 6= 0 on an open interval I that contains a
(except possibly at x = a). Suppose that
lim f (x) = 0 and lim g(x) = 0
x→a
x→a
or
lim f (x) = ±∞ and lim g(x) = ±∞
x→a
x→a
which gives us our two indeterminate forms
0
∞
or
. Then
0
∞
f (x)
f 0 (x)
lim
= lim 0
x→a g(x)
x→a g (x)
So what does this say?
It says the the limit of a quotient is equal to the limit of a quotient of their respected
derivatives.
105
3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
Warning! This is not the quotient rule! You are taking the derivatives of f (x) and g(x)
separately.
sin(x)
x→0
x
1. lim
Since this has the indeterminate form of type
0
, we can use L’Hospital’s Rule.
0
sin(x)
cos(x)
= lim
x→0
x→0
x
1
cos(0)
=
1
1
=
1
= 1
lim
ln x − 1
x→∞ x − 1
2. lim
This has the indterminate form of type
∞
, so we can use L’Hospital’s Rule.
∞
ln x − 1
=
x→∞ x − 1
lim
lim
1
x
1
1
= lim
x→∞ x
= 0
x→∞
e3t − 1
t→0
t2
3. lim
0
This has the indeterminate form of type . You can use the rule as many times as you
0
need as long as it has the appropriate indeterminate form.
106
3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
e3t − 1
3e3t
=
lim
t→0
t→0 2t
t2
∗∗∗
9e3t
= lim
t→0 2
9e3·0
=
2
= 9/2
lim
∗∗∗
Did you notice I used L’Hospitals Rule a second time? Was I allowed to?
3e0
3
= . This is not one of our
0
0
indeterminate types. In this case, it means there is an asymptote at t = 0. We need
The answer is no. If you plug in t = 0, you get
to look at the left and right hand limits.
lim−
t→0
3e3t
= −∞
2t
lim+
t→0
3e3t
=∞
2t
Now we know how the function behaves at the asymptote.
So what did we learn here? Using L’Hospitals Rule is great (when you can use it). If
you don’t get one of the indeterminate forms, you can’t apply the rule.
107
3.6 Indeterminate Forms and L’Hospital’s Rule
3.6.2
Brian E. Veitch
Indeterminate Products
Suppose limx→a f (x) = 0 and limx→a g(x) = ±∞. The limit limx→a f (x)g(x) is called an
indeterminate product.
We also call it Indeterminate Form of type 0 · ∞.
We have two ways of dealing with this type.
1. Rewrite f (x) · g(x) as
f (x)
.
1/g(x)
0
With this new form, we have an indeterminate form of type . We now use L’Hospital’s
0
Rule.
2. Rewrite f (x) · g(x) as
g(x)
1/f (x)
With this form, we have an indeterminate form of type
∞
. We now use L’Hopsital’s
∞
Rule.
Example 3.36.
1
1. lim x tan
x→∞
x
Note that x → ∞ and tan(1/x) → 0 as x → ∞.
We either write this as
tan(1/x)
x
. The first option looks promising.
or
1/x
(tan(1/x))−1
108
3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
tan(1/x)
x→∞
1/x
sec2 (1/x) · − x12
= lim
x→∞
− x12
lim x tan(1/x) =
x→∞
=
lim
lim sec2 (1/x)
x→∞
= sec2 (0)
= 1
2. lim xex
x→−∞
This has the indeterminate form of type −∞ · 0
We can either write xex as
x
ex
or −x
1/x
e
I’d like you to try to use the first one. You’ll find that the denominator won’t go away.
We are going to choose the second option.
x
x→−∞ e−x
1
= lim
x→−∞ −e−x
= lim −ex
lim xex =
x→−∞
lim
x→−∞
= 0
3.6.3
Indeterminate Differences
Suppose limx→a f (x) = ∞ and limx→a g(x) = ∞. Then the limit
109
3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
lim f (x) − g(x)
x→a
is called an Indeterminate Form of Type ∞ − ∞ . The doesn’t mean the limit will
be 0 or ∞. It is also possible that the limit is some finite number.
Example 3.37.
1. lim
x→∞
√
x2 + x − x
This has the indeterminate form of type ∞−∞. Our approach is to simplify somehow.
We can use common denominators, conjugates, factoring, pretty much anything that
gets us to one single fraction.
In this case, we’ll multiply top and bottom by the conjugate.
lim
x→∞
√
x2 + x − x =
=
=
=
2. lim+
x→0
1
1
− x
x e −1
√
x2 + x + x
lim x2 + x − x · √
x→∞
x2 + x + x
(x2 + x) − x2
lim √
x→∞
x2 + x + x
x
lim √
x→∞
x2 + x + x
x
lim q
x→∞ 1
1
1+ x +1
x
√
=
1
lim q
x→∞
1 + x1 + 1
=
1
2
110
3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
This has the indeterminate form of ∞ − ∞.
Again, we would like to make this into one fraction. That way, we can hope for
lim+
x→0
1
1
− x
x e −1
=
x→0
=
=
=
=
=
=
3.6.4
lim+
lim
x→0+
lim
x→0+
lim
x→0+
lim
x→0+
lim
x→0+
0
∞
or
0
∞
ex − 1
x
−
x
x
x(e − 1) x(e − 1)
ex − x − 1
x(ex − 1)
ex − 1
xex + (ex − 1)
ex
xex + ex + ex
ex
xex + 2ex
1
x+2
1
2
Indeterminate Powers
Consider the following limit
lim (f (x))g(x)
x→a
1. If limx→a f (x) = 0 and limx→a g(x) = 0, then we have an indeterminate form of type
00 .
2. If limx→a f (x) = ∞ and limx→a g(x) = 0, then we have an indeterminate form of type
∞0 .
3. If limx→a f (x) = 1 and limx→a g(x) = ±∞, then we have an indeterminate form of
type 1∞ .
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3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
Here’s how we approach these types. Instead of working with
y = (f (x))g(x)
we work on
ln y = g(x) · ln(f (x))
From here it’s likely we have one of our previous indeterminate forms.
Example 3.38.
1. lim x1/x
x→∞
We have the indeterminate form of type ∞0
1
Let ln y = ln x1/x = · ln(x)
x
Now let’s find the limit
1
ln(x)
x→∞ x
lim
1
ln(x)
x
ln(x)
= lim
x→∞
x
1/x
= lim
x→∞ 1
= 0
lim ln y =
x→∞
lim
x→∞
So as x → ∞, we have ln y → 0. If we exponentiate both sides
eln y → e0
y→1
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3.6 Indeterminate Forms and L’Hospital’s Rule
Brian E. Veitch
Therefore, lim x1/x = 1
x→∞
2. lim+ (1 + sin(4x))cot(x)
x→0
Now we have type 1∞
Instead of lim y = lim(1 + sin(4x))cot(x) , let’s work on
lim ln y = lim+ cot(x) ln(1 + sin(4x))
x→0+
x→0
lim+ cot(x) ln(1 + sin(4x)) =
x→0
=
=
lim+
x→0
lim
ln(1 + sin(4x))
tan(x)
1
· 4 cos(4x)
1+sin(4x)
x→0+
1
1+sin(0)
sec2 (x)
· 4 cos(0)
sec2 (0)
= 4
If ln y → 4 as x → 0+ , then y → e4 as x → 0+ .
Therefore,
lim (1 + sin(4x))cot(x) = e4
x→0+
113