In this video we learn how to solve limits by factoring and cancelling. This is one of the most simple and powerful techniques for solving limits. Watch video: http://www.youtube.com/watch?v=r0Qw5gZuTYE For more videos and lessons: http://www.intuitive-calculus.com/solving-limits.html

3. Example 1

4. Example 1
Let’s consider the limit:

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

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.

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!.

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:

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
=

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

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

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)

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 =

15. 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

16. Example 1

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

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

20. 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!

21. Example 2

22. Example 2
lim
x→1
x3 − 1
x − 1

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

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

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

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

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

30. 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

31. Example 3

32. Example 3
lim
h→0
(a + h)3 − a3
h

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

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
=

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

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

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

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

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

42. 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