In this presentation I present the concept of limits at infinity and the basic technique for solving them. We solve an example step by step.

3. What are Limits at Infinity?

4. What are Limits at Infinity?
Limits at infinity are just like normal limits.

5. What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.

6. What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
lim
x→∞
f (x)

7. What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
lim
x→∞
f (x)
This means that x is growing without bounds.

8. What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
lim
x→∞
f (x)
This means that x is growing without bounds.
Or that it takes values greater than any number you can come up
with.

9. The Basic Technique For Solving Limits at Infinity

10. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:

11. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.

12. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
For example:

13. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
For example:
2x2
− 1
x2 + x

14. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
For example:
2x2
− 1
x2 + x
The greatest exponent is 2.

15. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
For example:
2x2
− 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.

16. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
For example:
2x2
− 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.
Here n is the greatest exponent we found before.

17. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
For example:
2x2
− 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.
Here n is the greatest exponent we found before.
In our example:

18. The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
For example:
2x2
− 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.
Here n is the greatest exponent we found before.
In our example:
2x2
−1
x2
x2+x
x2

19. 3. Now we simplify our expression algebraically:

20. 3. Now we simplify our expression algebraically:
2x2−1
x2
x2+x
x2

21. 3. Now we simplify our expression algebraically:
2x2−1
x2
x2+x
x2
=
2x2
x2 − 1
x2
x2
x2 + x
x2

22. 3. Now we simplify our expression algebraically:
2x2−1
x2
x2+x
x2
=
2 x2
x2
− 1
x2
x2
x2 + x
x2

23. 3. Now we simplify our expression algebraically:
2x2−1
x2
x2+x
x2
=
2 x2
x2
− 1
x2
x2
x2
+ x
x2

24. 3. Now we simplify our expression algebraically:
2x2−1
x2
x2+x
x2
=
2 x2
x2
− 1
x2
x2
x2
+ ¡x
x£2

25. 3. Now we simplify our expression algebraically:
2x2−1
x2
x2+x
x2
=
2 x2
x2
− 1
x2
x2
x2
+ ¡x
x£2
=
2 − 1
x2
1 + 1
x

26. 3. Now we simplify our expression algebraically:
2x2−1
x2
x2+x
x2
=
2 x2
x2
− 1
x2
x2
x2
+ ¡x
x£2
=
2 − 1
x2
1 + 1
x
4. Now we calculate the limit directly.

27. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3

28. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.

29. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:

30. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:
lim
x→∞
4x3
x3 − 2x2
x3 + 1
x3
5x3
x3 − 3
x3

31. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:
lim
x→∞
4 x3
x3
− 2x2
x3 + 1
x3
5x3
x3 − 3
x3

32. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:
lim
x→∞
4 x3
x3
− 2 x2
x£3
+ 1
x3
5x3
x3 − 3
x3

33. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:
lim
x→∞
4 x3
x3
− 2 x2
x£3
+ 1
x3
5 x3
x3
− 3
x3

34. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:
lim
x→∞
4 x3
x3
− 2 x2
x£3
+ 1
x3
5 x3
x3
− 3
x3
And we’re left with:

35. The First Example
Let’s find the following limit:
lim
x→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:
lim
x→∞
4 x3
x3
− 2 x2
x£3
+ 1
x3
5 x3
x3
− 3
x3
And we’re left with:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3

36. The First Example
Now we can calculate the limit directly:

37. The First Example
Now we can calculate the limit directly:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3

38. The First Example
Now we can calculate the limit directly:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3
How do we do that?!

39. The First Example
Now we can calculate the limit directly:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3
How do we do that?!
Just observe that anything that is divided by x will go to zero.

40. The First Example
Now we can calculate the limit directly:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3
How do we do that?!
Just observe that anything that is divided by x will go to zero.
Why?

41. The First Example
Now we can calculate the limit directly:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3
How do we do that?!
Just observe that anything that is divided by x will go to zero.
Why?
Because x is approaching ∞!

42. The First Example
Now we can calculate the limit directly:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3
How do we do that?!
Just observe that anything that is divided by x will go to zero.
Why?
Because x is approaching ∞!
Just try with your calculator. Divide anything by a very big
number and you’ll get very close to 0.

43. The First Example

44. The First Example
So, using that fact, we have that:

45. The First Example
So, using that fact, we have that:
lim
x→∞
4 − 2
x + 1
x3
5 − 3
x3

46. The First Example
So, using that fact, we have that:
lim
x→∞
4 −
¡
¡!
0
2
x + 1
x3
5 − 3
x3

47. The First Example
So, using that fact, we have that:
lim
x→∞
4 −
¡
¡!
0
2
x +
U
0
1
x3
5 − 3
x3

48. The First Example
So, using that fact, we have that:
lim
x→∞
4 −
¡
¡!
0
2
x +
U
0
1
x3
5 −
U
0
3
x3

49. The First Example
So, using that fact, we have that:
lim
x→∞
4 −
¡
¡!
0
2
x +
U
0
1
x3
5 −
U
0
3
x3
And we’re left with:

50. The First Example
So, using that fact, we have that:
lim
x→∞
4 −
¡
¡!
0
2
x +
U
0
1
x3
5 −
U
0
3
x3
And we’re left with:
lim
x→∞
4
5

51. The First Example
So, using that fact, we have that:
lim
x→∞
4 −
¡
¡!
0
2
x +
U
0
1
x3
5 −
U
0
3
x3
And we’re left with:
lim
x→∞
4
5
=
4
5

52. The First Example
So, using that fact, we have that:
lim
x→∞
4 −
¡
¡!
0
2
x +
U
0
1
x3
5 −
U
0
3
x3
And we’re left with:
lim
x→∞
4
5
=
4
5
That’s the final answer!