In this video we solve two examples of limits at infinity: one that can be solved in two ways and the limit of an infinite sum.

3. A Problem That Can Be Solved In Two Ways
First Our Basic Technique

4. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x

5. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.

6. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent.

7. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1.

8. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:

9. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:
lim
x→∞
x
x + 1
x
x
x

10. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:
lim
x→∞
¡x
¡x
+ 1
x
x
x

11. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:
lim
x→∞
¡x
¡x
+ 1
x
¡x
¡x

12. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:
lim
x→∞
¡x
¡x
+ 1
x
¡x
¡x
= lim
x→∞
1 +
1
x

13. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:
lim
x→∞
¡x
¡x
+ 1
x
¡x
¡x
= lim
x→∞
1 +
1
x
This one is easy:

14. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:
lim
x→∞
¡x
¡x
+ 1
x
¡x
¡x
= lim
x→∞
1 +
1
x
This one is easy:
lim
x→∞
1 +
£
£
£#
0
1
x

15. A Problem That Can Be Solved In Two Ways
First Our Basic Technique
lim
x→∞
x + 1
x
Let’s try first our technique.
First, identify the greatest exponent. It is 1. Now, divide
everything by x:
lim
x→∞
¡x
¡x
+ 1
x
¡x
¡x
= lim
x→∞
1 +
1
x
This one is easy:
lim
x→∞
1 +
£
£
£#
0
1
x
= 1

16. A Problem That Can Be Solved In Two Ways
Now a Faster Method

17. A Problem That Can Be Solved In Two Ways
Now a Faster Method
Let’s try a different approach:

18. A Problem That Can Be Solved In Two Ways
Now a Faster Method
Let’s try a different approach:
lim
x→∞
x + 1
x

19. A Problem That Can Be Solved In Two Ways
Now a Faster Method
Let’s try a different approach:
lim
x→∞
x + 1
x
In this case we can directly separate the terms:

20. A Problem That Can Be Solved In Two Ways
Now a Faster Method
Let’s try a different approach:
lim
x→∞
x + 1
x
In this case we can directly separate the terms:
lim
x→∞
x
x
+
1
x

21. A Problem That Can Be Solved In Two Ways
Now a Faster Method
Let’s try a different approach:
lim
x→∞
x + 1
x
In this case we can directly separate the terms:
lim
x→∞
&x
&x
+
1
x

22. A Problem That Can Be Solved In Two Ways
Now a Faster Method
Let’s try a different approach:
lim
x→∞
x + 1
x
In this case we can directly separate the terms:
lim
x→∞
&x
&x
+
1
x
= lim
x→∞
1 +
1
x

23. A Problem That Can Be Solved In Two Ways
Now a Faster Method
Let’s try a different approach:
lim
x→∞
x + 1
x
In this case we can directly separate the terms:
lim
x→∞
&x
&x
+
1
x
= lim
x→∞
1 +
1
x
= 1

24. A Different Problem

25. A Different Problem
lim
n→∞
1 + 2 + 3 + ... + n
n2

26. A Different Problem
lim
n→∞
1 + 2 + 3 + ... + n
n2
As n → ∞, the numerator becomes an infinite sum.

27. A Different Problem
lim
n→∞
1 + 2 + 3 + ... + n
n2
As n → ∞, the numerator becomes an infinite sum.
The numerator is an arithmetic progresion.

28. A Different Problem

29. A Different Problem
We can find a formula for this sum:

30. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n

31. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1

32. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn =

33. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1)

34. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2)

35. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2) + (n − 2 + 3)

36. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2) + (n − 2 + 3) + ...

37. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2) + (n − 2 + 3) + ... + (n + 1)

38. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) +$$$$$$X(n + 1)
(n − 1 + 2) + (n − 2 + 3) + ... + (n + 1)

39. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) +$$$$$$X(n + 1)
(n − 1 + 2) +$$$$$$X(n + 1)
(n − 2 + 3) + ... + (n + 1)

40. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) +$$$$$$X(n + 1)
(n − 1 + 2) +$$$$$$X(n + 1)
(n − 2 + 3) + ... + (n + 1)
n terms

41. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn+Sn = (n + 1) +$$$$$$X(n + 1)
(n − 1 + 2) +$$$$$$X(n + 1)
(n − 2 + 3) + ... + (n + 1)
n terms
= n(n+1)

42. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn+Sn = (n + 1) +$$$$$$X(n + 1)
(n − 1 + 2) +$$$$$$X(n + 1)
(n − 2 + 3) + ... + (n + 1)
n terms
= n(n+1)
2Sn = n(n + 1)

43. A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn+Sn = (n + 1) +$$$$$$X(n + 1)
(n − 1 + 2) +$$$$$$X(n + 1)
(n − 2 + 3) + ... + (n + 1)
n terms
= n(n+1)
2Sn = n(n + 1)
Sn =
n(n + 1)
2

44. A Different Problem

45. A Different Problem
Now we can apply that formula to our problem:

46. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2

47. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2

48. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2

49. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2

50. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.

51. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:

52. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2 + n
n2
2n2
n2

53. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2
+ n
n2
2n2
n2

54. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2
+ ¡n
n£2
2n2
n2

55. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2
+ ¡n
n£2
2 n2
n2

56. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2
+ ¡n
n£2
2 n2
n2
= lim
n→∞
1 + 1
n
2

57. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2
+ ¡n
n£2
2 n2
n2
= lim
n→∞
1 +
¡
¡!
0
1
n
2

58. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2
+ ¡n
n£2
2 n2
n2
= lim
n→∞
1 +
¡
¡!
0
1
n
2
=
1
2

59. A Different Problem
Now we can apply that formula to our problem:
lim
n→∞
1 + 2 + 3 + ... + n
n2
lim
n→∞
n(n+1)
2
n2
= lim
n→∞
n(n + 1)
2n2
lim
n→∞
n2 + n
2n2
Now let’s apply our technique.
Let’s divide everything by n2:
lim
n→∞
n2
n2
+ ¡n
n£2
2 n2
n2
= lim
n→∞
1 +
¡
¡!
0
1
n
2
=
1
2