The document discusses L'Hopital's rule for evaluating ambiguous limit forms involving infinity and zero, namely ∞0, 00, and 1∞. These limits arise when evaluating limits of the form lim b(x)e(x), where b(x) and e(x) are functions approaching particular values. The document provides examples of using L'Hopital's rule to find the limit by first evaluating the limit of the log of the expression. One example calculates the limit lim x1/x by rewriting it as a log limit and applying L'Hopital's rule.

1. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

2. Theorem: Let k > 0, then lim k1/x = 1.
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

3. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K

4. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K K1/2

5. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K K1/2 K1/3

6. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K K1/2 K1/3 K1/4

7. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK K1/2 K1/3 K1/4

8. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K K1/2 K1/3 K1/4

9. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4

10. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what will
happen if k changes?

11. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what will
happen if k changes?
In general, if b(x)  ∞ and e(x)  0 as x  a, what
happens to b(x)e(x) as x  a?

12. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what will
happen if k changes?
In general, if b(x)  ∞ and e(x)  0 as x  a, what
happens to b(x)e(x) as x  a? We say these limit
problems are of the form ∞0.

13. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what will
happen if k changes?
In general, if b(x)  ∞ and e(x)  0 as x  a, what
happens to b(x)e(x) as x  a? We say these limit
problems are of the form ∞0.
Similarly, if b(x)  0 and e(x)  0 as x  a, we say
these problems "lim b(x)e(x)" are of the form 00.
xa

14. Theorem: Let k > 0, then lim k1/x = 1.
In picture:
x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what will
happen if k changes?
In general, if b(x)  ∞ and e(x)  0 as x  a, what
happens to b(x)e(x) as x  a? We say these limit
problems are of the form ∞0.
Similarly, if b(x)  0 and e(x)  0 as x  a, we say
these problems "lim b(x)e(x)" are of the form 00.
xa
Lastly, we have the 1∞ forms, i.e. the problems of
"lim b(x)e(x)" where b(x)  1 and e(x)  ∞.
xa

15. be a constant, then lim k1/x = 1.x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what will
happen if k changes?
In general, if b(x)  ∞ and e(x)  0 as x  a, what
happens to b(x)e(x) as x  a? We say these limit
problems are of the form ∞0.
In the last section we investigated the ambiguous
forms of "0/0 ", "∞/∞" and "0*∞" limit–problems.
x∞
The ambiguity of these forms is due to their fractional
format where the limits are determined by the rates
the numerators and denominators approach 0 (or ∞).

16. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
Theorem: Let k > 0, then lim k1/x = 1.
We outline the results and give examples below and
Let k > 0, then lim k1/x = 1.

17. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) . (∞0, 00, 1∞)

18. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
(∞0, 00, 1∞)

19. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x.
x∞
(∞0, 00, 1∞)

20. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
(∞0, 00, 1∞)

21. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
We find the limit of Ln(x1/x) = first.Ln(x)
x
(∞0, 00, 1∞)

22. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
(∞0, 00, 1∞)

23. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
this is ∞/∞ form so use
the L'Hopital's Rule.
(∞0, 00, 1∞)

24. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
=
this is ∞/∞ form so use
the L'Hopital's Rule.
lim [Ln(x)]'
[x]'x∞
(∞0, 00, 1∞)

25. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
=
this is ∞/∞ form so use
the L'Hopital's Rule.
lim [Ln(x)]'
[x]'x∞
= lim 1
xx∞
(∞0, 00, 1∞)

26. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
=
this is ∞/∞ form so use
the L'Hopital's Rule.
lim [Ln(x)]'
[x]'x∞
= lim 1
xx∞
= 0.
(∞0, 00, 1∞)

27. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
=
this is ∞/∞ form so use
the L'Hopital's Rule.
lim [Ln(x)]'
[x]'x∞
= lim 1
xx∞
= 0.
So lim x1/x =
x∞
lim eLn(x )1/x
(∞0, 00, 1∞)

28. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
=
this is ∞/∞ form so use
the L'Hopital's Rule.
lim [Ln(x)]'
[x]'x∞
= lim 1
xx∞
= 0.
So lim x1/x =
x∞
lim eLn(x )1/x
(∞0, 00, 1∞)

29. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
=
this is ∞/∞ form so use
the L'Hopital's Rule.
lim [Ln(x)]'
[x]'x∞
= lim 1
xx∞
= 0.
So lim x1/x =
x∞
lim eLn(x )1/x
(∞0, 00, 1∞)
0

30. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three ambiguousforms related to
limits of power functions b(x)e(x) .
We find their limits by first finding the limits of the
logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)
x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)
x
lim Ln(x)
xx∞ x∞
=
this is ∞/∞ form so use
the L'Hopital's Rule.
lim [Ln(x)]'
[x]'x∞
= lim 1
xx∞
= 0.
So lim x1/x =
x∞
lim eLn(x )1/x
= e0 = 1 .
0
(∞0, 00, 1∞)

31. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+

32. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+ x0+

33. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+ x0+
lim Ln(x)
x0+
x-1 =

34. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+ x0+
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2

35. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+ x0+
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2
–x

36. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+ x0+
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
–x

37. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
x0+
x0+
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
–x

38. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
x0+
x0+
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
–x

39. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
–x

40. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
–x

41. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x]limx∞
limx∞
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
–x

42. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x]limx∞
limx∞
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
–x
1

43. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
–x
1

44. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
=
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
–x
1

45. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= =
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

46. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= limx∞
[Ln(x + c) – Ln(x)]'
[x-1]'
=
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

47. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= limx∞
[Ln(x + c) – Ln(x)]'
[x-1]'
=
limx∞
1/(x + c) – 1/x
-1/x2
=
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

48. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= limx∞
[Ln(x + c) – Ln(x)]'
[x-1]'
=
limx∞
1/(x + c) – 1/x
-1/x2
= limx∞
x2 – x(x + c)
-(x + 1)
=
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

49. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= limx∞
[Ln(x + c) – Ln(x)]'
[x-1]'
=
limx∞
1/(x + c) – 1/x
-1/x2
= limx∞
x2 – x(x + c)
-(x + 1)
= limx∞
cx
x + 1
=
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

50. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= limx∞
[Ln(x + c) – Ln(x)]'
[x-1]'
=
limx∞
1/(x + c) – 1/x
-1/x2
= limx∞
x2 – x(x + c)
-(x + 1)
= limx∞
cx
x + 1
= = c
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

51. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= limx∞
[Ln(x + c) – Ln(x)]'
[x-1]'
=
limx∞
1/(x + c) – 1/x
-1/x2
= limx∞
x2 – x(x + c)
-(x + 1)
= limx∞
cx
x + 1
= = c
Therefore lim (1 + c/x)x = ec.x∞
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

52. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)
x0+
lim Ln(xx) =lim xLn(x) =x0+
So lim xx =
x0+
lim eLn(x )x
= 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)
x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)limx∞
limx∞
limx∞
Ln(x + c) – Ln(x)
x-1
= limx∞
[Ln(x + c) – Ln(x)]'
[x-1]'
=
limx∞
1/(x + c) – 1/x
-1/x2
= limx∞
x2 – x(x + c)
-(x + 1)
= limx∞
cx
x + 1
= = c
Therefore lim (1 + c/x)x = ec.x∞
lim Ln(x)
x0+
x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
0
L'Hopital
–x
1

53. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞.

54. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.

55. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1

56. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x

57. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x
( form, use
L'Hopital's rule)
0
0

58. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x
( form, use
L'Hopital's rule)
0
0
x0 [xsin(x)]'
[sin(x) – x]'
= lim

59. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x
( form, use
L'Hopital's rule)
0
0
x0 [xsin(x)]'
[sin(x) – x]'
= lim
x0 sin(x)+ xcos(x)
cos(x) – 1
= lim

60. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x
( form, use
L'Hopital's rule)
0
0
x0 [xsin(x)]'
[sin(x) – x]'
= lim
x0 sin(x)+ xcos(x)
cos(x) – 1
= lim ( form, use L'Hopital's again)
0
0

61. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x
( form, use
L'Hopital's rule)
0
0
x0 [xsin(x)]'
[sin(x) – x]'
= lim
x0 sin(x)+ xcos(x)
cos(x) – 1
= lim ( form, use L'Hopital's again)
0
0
x0 [sin(x)+ xcos(x)]'
[cos(x) – 1]'
= lim

62. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x
( form, use
L'Hopital's rule)
0
0
x0 [xsin(x)]'
[sin(x) – x]'
= lim
x0 sin(x)+ xcos(x)
cos(x) – 1
= lim ( form, use L'Hopital's again)
0
0
x0 [sin(x)+ xcos(x)]'
[cos(x) – 1]'
= lim
x0 cos(x) + cos(x) – xsin(x)
-sin(x)
= lim

63. D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the ambiguousforms are of the
form +∞ – ∞ or -∞ + ∞. In these cases, we put the
formulas into fractions first then take the limits.
1
sin(x)
1
lim( ) =x0 x
1
sin(x)
1 limx0 xsin(x)
sin(x) – x
( form, use
L'Hopital's rule)
0
0
x0 [xsin(x)]'
[sin(x) – x]'
= lim
x0 sin(x)+ xcos(x)
cos(x) – 1
= lim ( form, use L'Hopital's again)
0
0
x0 [sin(x)+ xcos(x)]'
[cos(x) – 1]'
= lim
x0 cos(x) + cos(x) – xsin(x)
-sin(x)
= lim = 0

64. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

65. L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)