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

Example 1

Example 1

Let’s consider the limit:

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

Example 1

Let’s consider the limit:

limx→2

x2 − 4

x − 2

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

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 !.

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:

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=

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

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

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

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)

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 =

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

Example 1

Example 1

What is the graph of this function?

Example 1

What is the graph of this function?

f (x) =x2 − 2

x + 2

Example 1

What is the graph of this function?

f (x) =x2 − 2

x + 2

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!

Example 2

Example 2

limx→1

x3 − 1

x − 1

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1=

Example 2

limx→1

x3 − 1

x − 1

Remember how to factor the numerator?

limx→1

x3 − 1

x − 1= lim

x→1

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

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

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

)

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

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

Example 3

Example 3

limh→0

(a+ h)3 − a3

h

Example 3

limh→0

(a+ h)3 − a3

h

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

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

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=

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

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

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

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

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

)

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

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

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