Today we focus on limits by factoring. We solve limits by factoring and cancelling. This is one of the basic techniques for solving limits. We talk about the idea behind this technique and we solve some examples step by step.

2. Example 1

3. Example 1
Let’s consider the limit:

4. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2

5. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.

6. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.

7. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:

8. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:
lim
x→2
x2 − 4
x − 2
=

9. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:
lim
x→2
x2 − 4
x − 2
= lim
x→2

10. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:
lim
x→2
x2 − 4
x − 2
= lim
x→2
(x − 2)(x + 2)
x − 2

11. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:
lim
x→2
x2 − 4
x − 2
= lim
x→2
$$$$(x − 2)(x + 2)
$$$x − 2

12. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:
lim
x→2
x2 − 4
x − 2
= lim
x→2
$$$$(x − 2)(x + 2)
$$$x − 2
= lim
x→2
(x + 2)

13. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:
lim
x→2
x2 − 4
x − 2
= lim
x→2
$$$$(x − 2)(x + 2)
$$$x − 2
= lim
x→2
(x + 2) = 2 + 2 =

14. Example 1
Let’s consider the limit:
lim
x→2
x2 − 4
x − 2
We note that at x = 2, our function is indeterminate.
It equals 0
0!.
But we can factor:
lim
x→2
x2 − 4
x − 2
= lim
x→2
$$$$(x − 2)(x + 2)
$$$x − 2
= lim
x→2
(x + 2) = 2 + 2 = 4

15. Example 1

16. Example 1
What is the graph of this function?

17. Example 1
What is the graph of this function?
f (x) =
x2 − 2
x + 2

18. Example 1
What is the graph of this function?
f (x) =
x2 − 2
x + 2

19. Example 1
What is the graph of this function?
f (x) =
x2 − 2
x + 2
It is the graph of x + 2, but with a hole!

20. Example 2

21. Example 2
lim
x→1
x3 − 1
x − 1

22. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?

23. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?
lim
x→1
x3 − 1
x − 1
=

24. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?
lim
x→1
x3 − 1
x − 1
= lim
x→1

25. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?
lim
x→1
x3 − 1
x − 1
= lim
x→1
(x − 1)(x2 + x + 1)
x − 1

26. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?
lim
x→1
x3 − 1
x − 1
= lim
x→1
$$$$(x − 1)(x2 + x + 1)
$$$x − 1

27. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?
lim
x→1
x3 − 1
x − 1
= lim
x→1
$$$$(x − 1)(x2 + x + 1)
$$$x − 1
= lim
x→1
x2
+ x + 1

28. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?
lim
x→1
x3 − 1
x − 1
= lim
x→1
$$$$(x − 1)(x2 + x + 1)
$$$x − 1
= lim
x→1
x2
+ x + 1 = 12
+ 1 + 1

29. Example 2
lim
x→1
x3 − 1
x − 1
Remember how to factor the numerator?
lim
x→1
x3 − 1
x − 1
= lim
x→1
$$$$(x − 1)(x2 + x + 1)
$$$x − 1
= lim
x→1
x2
+ x + 1 = 12
+ 1 + 1 = 3

30. Example 3

31. Example 3
lim
h→0
(a + h)3 − a3
h

32. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:

33. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h

34. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=

35. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=
lim
h→0
3a2h + 3ah2 + h3
h

36. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=
lim
h→0
3a2h + 3ah2 + h3
h
= lim
h→0

37. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=
lim
h→0
3a2h + 3ah2 + h3
h
= lim
h→0
h 3a2 + 3ah + h2
h

38. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=
lim
h→0
3a2h + 3ah2 + h3
h
= lim
h→0
¡h 3a2 + 3ah + h2
¡h

39. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=
lim
h→0
3a2h + 3ah2 + h3
h
= lim
h→0
¡h 3a2 + 3ah + h2
¡h
= lim
h→0
3a2
+ 3ah + h2

40. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=
lim
h→0
3a2h + 3ah2 + h3
h
= lim
h→0
¡h 3a2 + 3ah + h2
¡h
= lim
h→0
3a2
+ 3ah + h2
= 3a2
+ 3a.0 + 02

41. Example 3
lim
h→0
(a + h)3 − a3
h
Here a is a constant. Let’s expand (a + h)3:
lim
h→0
a3 + 3a2h + 3ah2 + h3 − a3
h
=
lim
h→0
3a2h + 3ah2 + h3
h
= lim
h→0
¡h 3a2 + 3ah + h2
¡h
= lim
h→0
3a2
+ 3ah + h2
= 3a2
+ 3a.0 + 02
= 3a2