limits by factoring
DESCRIPTION
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.htmlTRANSCRIPT
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