day 3 of free intuitive calculus course: limits by factoring

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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.

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Page 1: Day 3 of Free Intuitive Calculus Course: Limits by Factoring
Page 2: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 1

Page 3: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 1

Let’s consider the limit:

Page 4: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

Page 5: Day 3 of Free Intuitive Calculus Course: 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 6: Day 3 of Free Intuitive Calculus Course: 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 7: Day 3 of Free Intuitive Calculus Course: 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 8: Day 3 of Free Intuitive Calculus Course: 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 9: Day 3 of Free Intuitive Calculus Course: 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 10: Day 3 of Free Intuitive Calculus Course: 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 11: Day 3 of Free Intuitive Calculus Course: 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: Day 3 of Free Intuitive Calculus Course: 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 13: Day 3 of Free Intuitive Calculus Course: 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 14: Day 3 of Free Intuitive Calculus Course: 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 15: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 1

Page 16: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 1

What is the graph of this function?

Page 17: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 1

What is the graph of this function?

f (x) =x2 − 2

x + 2

Page 18: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 1

What is the graph of this function?

f (x) =x2 − 2

x + 2

Page 19: Day 3 of Free Intuitive Calculus Course: 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 20: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 2

Page 21: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Page 22: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

Page 23: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1=

Page 24: Day 3 of Free Intuitive Calculus Course: 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 25: Day 3 of Free Intuitive Calculus Course: 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 26: Day 3 of Free Intuitive Calculus Course: 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: Day 3 of Free Intuitive Calculus Course: 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 28: Day 3 of Free Intuitive Calculus Course: 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 29: Day 3 of Free Intuitive Calculus Course: 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 30: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 3

Page 31: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

Page 32: Day 3 of Free Intuitive Calculus Course: Limits by Factoring

Example 3

limh→0

(a+ h)3 − a3

h

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

Page 33: Day 3 of Free Intuitive Calculus Course: 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 34: Day 3 of Free Intuitive Calculus Course: 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: Day 3 of Free Intuitive Calculus Course: 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 36: Day 3 of Free Intuitive Calculus Course: 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 37: Day 3 of Free Intuitive Calculus Course: 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 38: Day 3 of Free Intuitive Calculus Course: 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: Day 3 of Free Intuitive Calculus Course: 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 40: Day 3 of Free Intuitive Calculus Course: 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 41: Day 3 of Free Intuitive Calculus Course: 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 42: Day 3 of Free Intuitive Calculus Course: Limits by Factoring