limits by factoring

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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

TRANSCRIPT

Page 1: Limits by Factoring
Page 2: Limits by Factoring
Page 3: Limits by Factoring

Example 1

Page 4: Limits by Factoring

Example 1

Let’s consider the limit:

Page 5: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

Page 6: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.

Page 7: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.

Page 8: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

Page 9: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

limx→2

x2 − 4

x − 2=

Page 10: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

limx→2

x2 − 4

x − 2= lim

x→2

Page 11: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

limx→2

x2 − 4

x − 2= lim

x→2

(x − 2)(x + 2)

x − 2

Page 12: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

limx→2

x2 − 4

x − 2= lim

x→2

����(x − 2)(x + 2)

���x − 2

Page 13: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

limx→2

x2 − 4

x − 2= lim

x→2

����(x − 2)(x + 2)

���x − 2

= limx→2

(x + 2)

Page 14: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

limx→2

x2 − 4

x − 2= lim

x→2

����(x − 2)(x + 2)

���x − 2

= limx→2

(x + 2) = 2 + 2 =

Page 15: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

We note that at x = 2, our function is indeterminate.It equals 0

0 !.But we can factor:

limx→2

x2 − 4

x − 2= lim

x→2

����(x − 2)(x + 2)

���x − 2

= limx→2

(x + 2) = 2 + 2 = 4

Page 16: Limits by Factoring

Example 1

Page 17: Limits by Factoring

Example 1

What is the graph of this function?

Page 18: Limits by Factoring

Example 1

What is the graph of this function?

f (x) =x2 − 2

x + 2

Page 19: Limits by Factoring

Example 1

What is the graph of this function?

f (x) =x2 − 2

x + 2

Page 20: Limits by Factoring

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!

Page 21: Limits by Factoring

Example 2

Page 22: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Page 23: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

Page 24: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1=

Page 25: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1= lim

x→1

Page 26: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1= lim

x→1

(x − 1)(x2 + x + 1)

x − 1

Page 27: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1= lim

x→1

����(x − 1)(x2 + x + 1)

���x − 1

Page 28: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1= lim

x→1

����(x − 1)(x2 + x + 1)

���x − 1

= limx→1

(x2 + x + 1

)

Page 29: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1= lim

x→1

����(x − 1)(x2 + x + 1)

���x − 1

= limx→1

(x2 + x + 1

)= 12 + 1 + 1

Page 30: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1= lim

x→1

����(x − 1)(x2 + x + 1)

���x − 1

= limx→1

(x2 + x + 1

)= 12 + 1 + 1 = 3

Page 31: Limits by Factoring

Example 3

Page 32: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Page 33: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

Page 34: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

a3 + 3a2h + 3ah2 + h3 − a3

h

Page 35: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

Page 36: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

limh→0

3a2h + 3ah2 + h3

h

Page 37: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

limh→0

3a2h + 3ah2 + h3

h= lim

h→0

Page 38: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

limh→0

3a2h + 3ah2 + h3

h= lim

h→0

h(3a2 + 3ah + h2

)h

Page 39: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

limh→0

3a2h + 3ah2 + h3

h= lim

h→0

�h(3a2 + 3ah + h2

)�h

Page 40: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

limh→0

3a2h + 3ah2 + h3

h= lim

h→0

�h(3a2 + 3ah + h2

)�h

= limh→0

(3a2 + 3ah + h2

)

Page 41: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

limh→0

3a2h + 3ah2 + h3

h= lim

h→0

�h(3a2 + 3ah + h2

)�h

= limh→0

(3a2 + 3ah + h2

)= 3a2 + 3a.0 + 02

Page 42: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Here a is a constant. Let’s expand (a+ h)3:

limh→0

��a3 + 3a2h + 3ah2 + h3 −��a3

h=

limh→0

3a2h + 3ah2 + h3

h= lim

h→0

�h(3a2 + 3ah + h2

)�h

= limh→0

(3a2 + 3ah + h2

)= 3a2 + 3a.0 + 02 = 3a2

Page 43: Limits by Factoring