L'Hopital Rule I

1. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)

2. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Let A = 2x and B = x2, as x → 1, we have that
lim A → 2 and lim B →1,

3. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Let A = 2x and B = x2, as x → 1, we have that
lim A → 2 and lim B →1,
so the function A + B = 2x + x2 → 2 + 1 = 3.

4. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Let A = 2x and B = x2, as x → 1, we have that
lim A → 2 and lim B →1,
so the function A + B = 2x + x2 → 2 + 1 = 3.
However, as x → 0, we have that
lim A → 0 and lim B → 0,
plugging them in the function A/B = 2x/x2 yields “0/0”
which is undefined.

5. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Let A = 2x and B = x2, as x → 1, we have that
lim A → 2 and lim B →1,
so the function A + B = 2x + x2 → 2 + 1 = 3.
However, as x → 0, we have that
lim A → 0 and lim B → 0,
plugging them in the function A/B = 2x/x2 yields “0/0”
which is undefined. Only after simplifying 2x/x2 as 2/x,
we see that limit A/B → ∞ as x → 0.

6. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Let A = 2x and B = x2, as x → 1, we have that
lim A → 2 and lim B →1,
so the function A + B = 2x + x2 → 2 + 1 = 3.
However, as x → 0, we have that
lim A → 0 and lim B → 0,
plugging them in the function A/B = 2x/x2 yields “0/0”
which is undefined. Only after simplifying 2x/x2 as 2/x,
we see that limit A/B → ∞ as x → 0.
The reciprocal B/A = x2/2x also yields “0/0” as x → 0,
simplifying x2/2x as x/2, we see that B/A → 0 as x → 0.

7. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
So “0/0” is an ambiguous form because this form may
yield different answers. Meaning that if we have “0/0”,
STOP! we need to reformulate the problem.
Let A = 2x and B = x2, as x → 1, we have that
lim A → 2 and lim B →1,
so the function A + B = 2x + x2 → 2 + 1 = 3.
However, as x → 0, we have that
lim A → 0 and lim B → 0,
plugging them in the function A/B = 2x/x2 yields “0/0”
which is undefined. Only after simplifying 2x/x2 as 2/x,
we see that limit A/B → ∞ as x → 0.
The reciprocal B/A = x2/2x also yields “0/0” as x → 0,
simplifying x2/2x as x/2, we see that B/A → 0 as x → 0.

8. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞,

9. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞, then the limits of A/B and AB
yield many ambiguous forms (with the inputs 0 and ∞.)

10. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Assuming A, B ≥ 0, the following are all the possibilities
if we plug 0 or ∞ as the limits of A and B:
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞, then the limits of A/B and AB
yield many ambiguous forms (with the inputs 0 and ∞.)

11. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Assuming A, B ≥ 0, the following are all the possibilities
if we plug 0 or ∞ as the limits of A and B:
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞, then the limits of A/B and AB
yield many ambiguous forms (with the inputs 0 and ∞.)
A/B: 0/0, ∞/∞, ∞/0, 0/∞, AB: ∞0, 00, ∞∞, 0∞

12. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Assuming A, B ≥ 0, the following are all the possibilities
if we plug 0 or ∞ as the limits of A and B:
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞, then the limits of A/B and AB
yield many ambiguous forms (with the inputs 0 and ∞.)
A/B: 0/0, ∞/∞, ∞/0, 0/∞, AB: ∞0, 00, ∞∞, 0∞

13. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Assuming A, B ≥ 0, the following are all the possibilities
if we plug 0 or ∞ as the limits of A and B:
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞, then the limits of A/B and AB
yield many ambiguous forms (with the inputs 0 and ∞.)
A/B: 0/0, ∞/∞, ∞/0, 0/∞, AB: ∞0, 00, ∞∞, 0∞
The cases ∞/0, 0/∞, ∞∞, 0∞ in the above forms are
determined since ∞/0 = ∞, ∞∞ = ∞, 0/∞ = 0, and 0∞ = 0.

14. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Assuming A, B ≥ 0, the following are all the possibilities
if we plug 0 or ∞ as the limits of A and B:
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞, then the limits of A/B and AB
yield many ambiguous forms (with the inputs 0 and ∞.)
A/B: 0/0, ∞/∞, ∞/0, 0/∞, AB: ∞0, 00, ∞∞, 0∞
The cases ∞/0, 0/∞, ∞∞, 0∞ in the above forms are
determined since ∞/0 = ∞, ∞∞ = ∞, 0/∞ = 0, and 0∞ = 0.
However, the forms 0/0, ∞/∞, ∞0, 00 above are
ambiguous.

15. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
Assuming A, B ≥ 0, the following are all the possibilities
if we plug 0 or ∞ as the limits of A and B:
In general, let A(x) and B(x) be two formulas such that
lim A and lim B → 0 or ∞, then the limits of A/B and AB
yield many ambiguous forms (with the inputs 0 and ∞.)
A/B: 0/0, ∞/∞, ∞/0, 0/∞, AB: ∞0, 00, ∞∞, 0∞
The cases ∞/0, 0/∞, ∞∞, 0∞ in the above forms are
determined since ∞/0 = ∞, ∞∞ = ∞, 0/∞ = 0, and 0∞ = 0.
However, the forms 0/0, ∞/∞, ∞0, 00 above are
ambiguous. We use the sequences of
n = 1, 2, ..→ ∞ and 1/n = 1, ½, 1/3, 1/4,.. → 0 to makes
examples to see why the forms ∞0, 00 are ambiguous.

16. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
n

17. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
01/n =
(1/n)0 =00

18. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 =00

19. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00

20. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00
Hence 00 is ambiguous.

21. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
The Ambiguous ∞0 form
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00
Hence 00 is ambiguous.

22. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
The Ambiguous ∞0 form
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00
Hence 00 is ambiguous.
(nn) 1/n =
∞0
(n)1/n =

23. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
The Ambiguous ∞0 form
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00
Hence 00 is ambiguous.
(nn) 1/n = √22, √33, √44, .. or 2, 3, 4, → ∞,
3 4
∞0
(n)1/n =

24. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
The Ambiguous ∞0 form
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00
Hence 00 is ambiguous.
(nn) 1/n = √22, √33, √44, .. or 2, 3, 4, → ∞,
3 4
∞0
(n)1/n = √2, √3, √4, .. → 1 (later on this)
3 4

25. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
The Ambiguous ∞0 form
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00
Hence 00 is ambiguous.
(nn) 1/n = √22, √33, √44, .. or 2, 3, 4, → ∞,
3 4
∞0
(n)1/n = √2, √3, √4, .. → 1 (later on this)
Hence ∞0 is ambiguous.
43

26. The Ambiguous Form of Limits
(the 0/0, ∞/∞, ∞*0, ∞0, 00, 1∞, ∞ – ∞ forms)
The expressions A1/n = √A as 1/n → 0 means that
we are taking higher and higher order of roots .
The Ambiguous 00 form
n
The Ambiguous ∞0 form
01/n = √0, √0, √0,.. → 0,
3 4
(1/n)0 = 1, 1, 1,.. → 1,00
Hence 00 is ambiguous.
(nn) 1/n = √22, √33, √44, .. or 2, 3, 4, → ∞,
3 4
∞0
(n)1/n = √2, √3, √4, .. → 1 (later on this)
Hence ∞0 is ambiguous.
Lastly we note that the form “∞ – ∞“ is the only
ambiguous case for adding or subtracting ∞’s and 0’s.
43

27. L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

28. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

29. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
We call this type of limit the " 0 "
0
ambiguous type.

30. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
" 0 "
0
form to the calculation of derivatives.
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
We call this type of limit the " 0 "
0
L'Hopital's rule passes the calculation of the
ambiguous type.

31. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
" 0 "
0
form to the calculation of derivatives.
Example: A. Find lim
sin(x)
xx0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
We call this type of limit the " 0 "
0
L'Hopital's rule passes the calculation of the
ambiguous type.

32. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
" 0 "
0
form to the calculation of derivatives.
Example: A. Find lim
sin(x)
xx0
Since lim sin(x) = 0 and lim x = 0,
x0 x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
We call this type of limit the " 0 "
0
L'Hopital's rule passes the calculation of the
ambiguous type.

33. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
" 0 "
0
form to the calculation of derivatives.
Example: A. Find lim
sin(x)
xx0
Since lim sin(x) = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
We call this type of limit the " 0 "
0
L'Hopital's rule passes the calculation of the
ambiguous type.

34. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
" 0 "
0
form to the calculation of derivatives.
Example: A. Find lim
sin(x)
xx0
Since lim sin(x) = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim
sin(x)
x = lim
[sin(x)]'
[x]'x0 x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
We call this type of limit the " 0 "
0
L'Hopital's rule passes the calculation of the
ambiguous type.

35. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
" 0 "
0
form to the calculation of derivatives.
Example: A. Find lim
sin(x)
xx0
Since lim sin(x) = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim
sin(x)
x = lim
[sin(x)]'
[x]'x0 x0
= lim
cos(x)
1x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
We call this type of limit the " 0 "
0
L'Hopital's rule passes the calculation of the
ambiguous type.

36. L'Hopital's Rule: Given lim f(x) = lim g(x) = 0, then
then lim f(x)
g(x)
= lim f '(x)
g'(x)
if the limit exists.
xa xa
xa xa
We call this type of limit the " 0 "
0
L'Hopital's rule passes the calculation of the
" 0 "
0
form to the calculation of derivatives.
Example: A. Find lim
sin(x)
xx0
Since lim sin(x) = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim
sin(x)
x = lim
[sin(x)]'
[x]'x0 x0
= lim
cos(x)
1x0
= 1.
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
ambiguous type.

37. B. Find lim ex – 1
xx0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

38. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

39. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

40. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
= lim
ex
1x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

41. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
= lim
ex
1x0
= 1.
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

42. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
= lim
ex
1x0
= 1.
C. Find lim ex – 1
x2x0+
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

43. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
= lim
ex
1x0
= 1.
C. Find lim ex – 1
x2x0+
Since lim ex – 1 = 0 and lim x2 = 0, use the L'Hopital's
rule:
x0+ x0+
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

44. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
= lim
ex
1x0
= 1.
C. Find lim ex – 1
x2x0+
Since lim ex – 1 = 0 and lim x2 = 0, use the L'Hopital's
rule: lim ex – 1
x2 = lim [ex – 1 ]'
[x2]'x0+
x0+ x0+
x0+
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

45. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
= lim
ex
1x0
= 1.
C. Find lim ex – 1
x2x0+
Since lim ex – 1 = 0 and lim x2 = 0, use the L'Hopital's
rule: lim ex – 1
x2 = lim [ex – 1 ]'
[x2]'x0+
= lim
ex
2xx0+
x0+ x0+
x0+
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

46. B. Find lim ex – 1
xx0
Since lim ex – 1 = 0 and lim x = 0, use the L'Hopital's
rule:
x0 x0
lim ex – 1
x = lim [ex – 1 ]'
[x]'x0 x0
= lim
ex
1x0
= 1.
C. Find lim ex – 1
x2x0+
Since lim ex – 1 = 0 and lim x2 = 0, use the L'Hopital's
rule: lim ex – 1
x2 = lim [ex – 1 ]'
[x2]'x0+
= lim
ex
2xx0+
= ∞.
x0+ x0+
x0+
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

47. D. Find lim 1
x(sin(1/x2))x+∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

48. D. Find lim 1
x(sin(1/x2))x+∞
Put the limit in the form of .
1/x
sin(1/x2)
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

49. D. Find lim 1
x(sin(1/x2))x+∞
Put the limit in the form of .
lim
1/x
x+∞
1/x = lim
x+∞
sin(1/x2) = 0, use the L'Hopital's rule:
sin(1/x2)
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

50. D. Find lim 1
x(sin(1/x2))x+∞
Put the limit in the form of .
lim
1/x
x+∞
1/x = lim
x+∞
sin(1/x2) = 0, use the L'Hopital's rule:
lim 1
x(sin(1/x2))x+∞
sin(1/x2)
= lim 1/x
sin(1/x2)x+∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

51. D. Find lim 1
x(sin(1/x2))x+∞
Put the limit in the form of .
lim
1/x
x+∞
1/x = lim
x+∞
sin(1/x2) = 0, use the L'Hopital's rule:
lim 1
x(sin(1/x2))x+∞
sin(1/x2)
= lim 1/x
sin(1/x2)x+∞
= lim
[1/x]'
[sin(1/x2)]'x+∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

52. D. Find lim 1
x(sin(1/x2))x+∞
Put the limit in the form of .
lim
1/x
x+∞
1/x = lim
x+∞
sin(1/x2) = 0, use the L'Hopital's rule:
lim 1
x(sin(1/x2))x+∞
sin(1/x2)
= lim 1/x
sin(1/x2)x+∞
= lim
[1/x]'
[sin(1/x2)]'x+∞
= lim
-x-2
cos(1/x2)(-2x-3)x+∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

53. D. Find lim 1
x(sin(1/x2))x+∞
Put the limit in the form of .
lim
1/x
x+∞
1/x = lim
x+∞
sin(1/x2) = 0, use the L'Hopital's rule:
lim 1
x(sin(1/x2))x+∞
sin(1/x2)
= lim 1/x
sin(1/x2)x+∞
= lim
[1/x]'
[sin(1/x2)]'x+∞
= lim
-x-2
cos(1/x2)(-2x-3)x+∞
= lim x
2cos(1/x2)x+∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

54. D. Find lim 1
x(sin(1/x2))x+∞
Put the limit in the form of .
lim
1/x
x+∞
1/x = lim
x+∞
sin(1/x2) = 0, use the L'Hopital's rule:
lim 1
x(sin(1/x2))x+∞
sin(1/x2)
= lim 1/x
sin(1/x2)x+∞
= lim
[1/x]'
[sin(1/x2)]'x+∞
= lim
-x-2
cos(1/x2)(-2x-3)x+∞
= lim x
2cos(1/x2)x+∞
= ∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

55. Remark: We need lim f(x) = lim g(x) = 0 for
L'Hopital's Rule (for the 0/0-form).
xa xa
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

56. Remark: We need lim f(x) = lim g(x) = 0 for
L'Hopital's Rule (for the 0/0-form).
xa xa
E. lim 2x
3x + 2x0
Example:
= 0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

57. Remark: We need lim f(x) = lim g(x) = 0 for
L'Hopital's Rule (for the 0/0-form).
xa xa
E. lim 2x
3x + 2x0
Example:
= 0 = lim [2x]'
[3x + 2]'x0
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

58. Remark: We need lim f(x) = lim g(x) = 0 for
L'Hopital's Rule (for the 0/0-form).
xa xa
E. lim 2x
3x + 2x0
Example:
= 0 = lim [2x]'
[3x + 2]'x0 = 2
3 .
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

59. Remark: We need lim f(x) = lim g(x) = 0 for
L'Hopital's Rule (for the 0/0-form).
xa xa
E. lim 2x
3x + 2x0
Example:
= 0 = lim [2x]'
[3x + 2]'x0 = 2
3 .
are of the
" ∞ " ambiguous forms.∞
If lim f(x) = lim g(x) = ∞, then limxa xa
f(x)
g(x)xa
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

60. Remark: We need lim f(x) = lim g(x) = 0 for
L'Hopital's Rule (for the 0/0-form).
xa xa
E. lim 2x
3x + 2x0
Example:
= 0 = lim [2x]'
[3x + 2]'x0 = 2
3 .
are of the
" ∞ " ambiguous forms. L'Hopital’s Rule applies∞
If lim f(x) = lim g(x) = ∞, then limxa xa
f(x)
g(x)xa
in these situations also,
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

61. Remark: We need lim f(x) = lim g(x) = 0 for
L'Hopital's Rule (for the 0/0-form).
xa xa
E. lim 2x
3x + 2x0
Example:
= 0 = lim [2x]'
[3x + 2]'x0 = 2
3 .
are of the
" ∞ " ambiguous forms. L'Hopital’s Rule applies∞
If lim f(x) = lim g(x) = ∞, then limxa xa
f(x)
g(x)xa
in these situations also,
lim f(x)
g(x)
= lim f '(x)
g'(x).xa xa
that is
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

62. Example: F. Find lim
3x2 – 4
2x2 + x – 5x∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

63. Example: F. Find lim
3x2 – 4
2x2 + x – 5x∞
It’s the form." ∞ "
∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

64. Example: F. Find lim
3x2 – 4
2x2 + x – 5x∞
It’s the form." ∞ "
∞
Hence 3x2 – 4
2x2 + x – 5
limx∞
=
[3x2 – 4]'
[2x2 + x – 5]'
limx∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

65. Example: F. Find lim
3x2 – 4
2x2 + x – 5x∞
It’s the form." ∞ "
∞
Hence 3x2 – 4
2x2 + x – 5
limx∞
=
[3x2 – 4]'
[2x2 + x – 5]'
limx∞
6x
4x + 1
limx∞
=
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

66. Example: F. Find lim
3x2 – 4
2x2 + x – 5x∞
It’s the form." ∞ "
∞
Hence 3x2 – 4
2x2 + x – 5
limx∞
=
[3x2 – 4]'
[2x2 + x – 5]'
limx∞
6x
4x + 1
limx∞
= use L'Hopital's Rule again
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

67. Example: F. Find lim
3x2 – 4
2x2 + x – 5x∞
It’s the form." ∞ "
∞
Hence 3x2 – 4
2x2 + x – 5
limx∞
=
[3x2 – 4]'
[2x2 + x – 5]'
limx∞
6x
4x + 1
limx∞
=
= use L'Hopital's Rule again
[6x]'
[4x + 1]'
limx∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

68. Example: F. Find lim
3x2 – 4
2x2 + x – 5x∞
It’s the form." ∞ "
∞
Hence 3x2 – 4
2x2 + x – 5
limx∞
=
[3x2 – 4]'
[2x2 + x – 5]'
limx∞
6x
4x + 1
limx∞
=
= use L'Hopital's Rule again
[6x]'
[4x + 1]'
limx∞
= 6
4
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
= 3
2

69. G. ex
P(x)x∞
where P(x) is a polynomial withFind lim
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
degree P(x) > 0.

70. G. ex
P(x)x∞
It’s the form." ∞ "
∞
where P(x) is a polynomial withFind lim
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
degree P(x) > 0.

71. G. ex
P(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where P(x) is a polynomial withFind lim
limx∞
ex
P(x)
[ex]'
[P(x)]'
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
degree P(x) > 0.

72. G. ex
P(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where P(x) is a polynomial withFind lim
limx∞
ex
P(x)
[ex]'
[P(x)]'
= limx∞
ex
[P(x)]'
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
degree P(x) > 0.

73. G. ex
P(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where P(x) is a polynomial withFind lim
limx∞
ex
P(x)
[ex]'
[P(x)]'
= limx∞
ex
[P(x)]'
Use the L'Hopital's Rule repeatedly if necessary,
eventually the denominator is a non-zero constant K.
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
degree P(x) > 0.

74. G. ex
P(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where P(x) is a polynomial withFind lim
limx∞
= limx∞
ex
K
= ∞
ex
P(x)
[ex]'
[P(x)]'
= limx∞
ex
[P(x)]'
Use the L'Hopital's Rule repeatedly if necessary,
eventually the denominator is a non-zero constant K.
Hence limx∞
ex
P(x)
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
degree P(x) > 0.

75. G. ex
P(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where P(x) is a polynomial withFind lim
limx∞
= limx∞
ex
K
We say that ex goes to ∞ faster than any polynomial.
= ∞
ex
P(x)
[ex]'
[P(x)]'
= limx∞
ex
[P(x)]'
Use the L'Hopital's Rule repeatedly if necessary,
eventually the denominator is a non-zero constant K.
Hence limx∞
ex
P(x)
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
degree P(x) > 0.

76. H. xp
Ln(x)x∞
where p > 0.Find lim
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

77. H. xp
Ln(x)x∞
It’s the form." ∞ "
∞
where p > 0.Find lim
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

78. H. xp
Ln(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where p > 0.Find lim
limx∞
xp
Ln(x)
[xp]'
[Ln(x)]'
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

79. H. xp
Ln(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where p > 0.Find lim
limx∞
= limx∞
pxp-1
x-1
xp
Ln(x)
[xp]'
[Ln(x)]'
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

80. H. xp
Ln(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where p > 0.Find lim
limx∞
= limx∞
pxp-1
x-1
xp
Ln(x)
[xp]'
[Ln(x)]'
= limx∞
pxp
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

81. H. xp
Ln(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where p > 0.Find lim
limx∞
= limx∞
pxp-1
x-1
xp
Ln(x)
[xp]'
[Ln(x)]'
= limx∞
pxp = ∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

82. H. xp
Ln(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where p > 0.Find lim
limx∞
= limx∞
pxp-1
x-1
We say that xp (p > 0) goes to ∞ faster than Ln(x).
xp
Ln(x)
[xp]'
[Ln(x)]'
= limx∞
pxp = ∞
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

83. H. xp
Ln(x)x∞
It’s the form." ∞ "
∞
Hence limx∞ =
where p > 0.Find lim
limx∞
= limx∞
pxp-1
x-1
We say that xp (p > 0) goes to ∞ faster than Ln(x).
xp
Ln(x)
[xp]'
[Ln(x)]'
= limx∞
pxp = ∞
The ∞/∞ form is the same as ∞*0 form since
∞/∞ = ∞ * (1/∞) and we may view that 1/∞ as 0.
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

84. I.
x0+
Find lim sin(x)Ln(x)
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

85. I.
x0+
Find lim sin(x)Ln(x)
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

86. I.
x0+
Find lim sin(x)Ln(x)
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
1/sin(x)
Ln(x)
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

87. I.
x0+
in the form.∞
∞
Find lim sin(x)Ln(x)
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
1/sin(x)
Ln(x)
=
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

88. I.
x0+
in the form.∞
∞
Hence lim sin(x)Ln(x) = lim
Find lim sin(x)Ln(x)
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
csc(x)
Ln(x)
x0+ x0+
1/sin(x)
Ln(x)
=
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

89. I.
x0+
in the form.∞
∞
Hence lim sin(x)Ln(x) = lim
=
Find lim sin(x)Ln(x)
lim
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
csc(x)
Ln(x)
x0+ x0+
[csc(x)]'
[Ln(x)]'
x0+
=
1/sin(x)
Ln(x)
=
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

90. I.
x0+
in the form.∞
∞
Hence lim sin(x)Ln(x) = lim
=
Find lim sin(x)Ln(x)
lim
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
csc(x)
Ln(x)
x0+ x0+
[csc(x)]'
[Ln(x)]'
x0+
= lim
-csc(x)cot(x)
1/x
x0+
1/sin(x)
Ln(x)
=
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

91. I.
x0+
in the form.∞
∞
Hence lim sin(x)Ln(x) = lim
=
Find lim sin(x)Ln(x)
lim
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
csc(x)
Ln(x)
x0+ x0+
[csc(x)]'
[Ln(x)]'
x0+
= lim
-csc(x)cot(x)
1/x
x0+
= lim
x
- sin(x)tan(x)
x0+
1/sin(x)
Ln(x)
=
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)

92. I.
x0+
in the form.∞
∞
Hence lim sin(x)Ln(x) = lim
=
Find lim sin(x)Ln(x)
lim
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
csc(x)
Ln(x)
x0+ x0+
[csc(x)]'
[Ln(x)]'
x0+
= lim
-csc(x)cot(x)
1/x
x0+
= lim
x
- sin(x)tan(x)
x0+
=
x
-sin(x)lim
x0+
tan(x)lim
x0+
1/sin(x)
Ln(x)
=

93. I.
x0+
in the form.∞
∞
Hence lim sin(x)Ln(x) = lim
=
Find lim sin(x)Ln(x)
lim
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
csc(x)
Ln(x)
x0+ x0+
[csc(x)]'
[Ln(x)]'
x0+
= lim
-csc(x)cot(x)
1/x
x0+
= lim
x
- sin(x)tan(x)
x0+
=
x
-sin(x)lim
x0+
tan(x)lim
x0+
0-1
1/sin(x)
Ln(x)
=

94. I.
x0+
in the form.∞
∞
Hence lim sin(x)Ln(x) = lim
=
Find lim sin(x)Ln(x)
lim
L'Hopital's Rule (0/0, ∞/∞, ∞*0 forms)
Lim sin(x) = 0 and lim Ln(x) = -∞ so its the 0*∞ form.
x0+ x0+
Write it as
csc(x)
Ln(x)
csc(x)
Ln(x)
x0+ x0+
[csc(x)]'
[Ln(x)]'
x0+
= lim
-csc(x)cot(x)
1/x
x0+
= lim
x
- sin(x)tan(x)
x0+
=
x
-sin(x)lim
x0+
tan(x) = 0lim
x0+
0-1
1/sin(x)
Ln(x)
=