limits by rationalization
DESCRIPTION
We solve limits by rationalizing. This is the second technique you may learn after limits by factoring. We solve two examples step by step. Watch video: http://www.youtube.com/watch?v=8CtpuojMJzA More videos and lessons: http://www.intuitive-calculus.com/solving-limits.htmlTRANSCRIPT
Example 1
Example 1
Let’s try to find the limit:
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h=
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.���
����*1√
a + h +√a√
a + h +√a
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a)
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
a + h − a
h(√
a + h +√a)
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
�a + h − �ah(√
a + h +√a)
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
�a + h − �ah(√
a + h +√a)
= limh→0
h
h(√
a + h +√a)
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
�a + h − �ah(√
a + h +√a)
= limh→0
�h
�h(√
a + h +√a)
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
�a + h − �ah(√
a + h +√a)
= limh→0
�h
�h(√
a + h +√a) = lim
h→0
1√a + h +
√a
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
�a + h − �ah(√
a + h +√a)
= limh→0
�h
�h(√
a + h +√a) = lim
h→0
1
����:
√a√
a + h +√a
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
�a + h − �ah(√
a + h +√a)
= limh→0
�h
�h(√
a + h +√a) = lim
h→0
1
����:
√a√
a + h +√a
=1
2√a
Example 1
Let’s try to find the limit:
limh→0
√a + h −
√a
h
We rationalize the expression.In this case by multiplying and dividing by the conjugate.
limh→0
√a + h −
√a
h= lim
h→0
√a + h −
√a
h.
√a + h +
√a√
a + h +√a
= limh→0
(√a + h
)2 − (√a)2
h(√
a + h +√a) = lim
h→0
�a + h − �ah(√
a + h +√a)
= limh→0
�h
�h(√
a + h +√a) = lim
h→0
1
����:
√a√
a + h +√a
=1
2√a
Example 2
Example 2
limx→0
√1 + x − 1
x
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x=
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.��
�����*1√
1 + x + 1√1 + x + 1
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1)
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
1 + x − 1
x(√
1 + x + 1)
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
�1 + x − �1x(√
1 + x + 1)
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
�1 + x − �1x(√
1 + x + 1)
= limx→0
x
x(√
1 + x + 1)
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
�1 + x − �1x(√
1 + x + 1)
= limx→0
�x
�x(√
1 + x + 1)
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
�1 + x − �1x(√
1 + x + 1)
= limx→0
�x
�x(√
1 + x + 1) = lim
x→0
1√1 + x + 1
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
�1 + x − �1x(√
1 + x + 1)
= limx→0
�x
�x(√
1 + x + 1) = lim
x→0
1
�����:
√1√
1 + x + 1
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
�1 + x − �1x(√
1 + x + 1)
= limx→0
�x
�x(√
1 + x + 1) = lim
x→0
1
�����:
√1√
1 + x + 1
=1
2
Example 2
limx→0
√1 + x − 1
x
We rationalize using the same method.
limx→0
√1 + x − 1
x= lim
x→0
√1 + x − 1
x.
√1 + x + 1√1 + x + 1
= limx→0
(√1 + x
)2 − 12
x(√
1 + x + 1) = lim
x→0
�1 + x − �1x(√
1 + x + 1)
= limx→0
�x
�x(√
1 + x + 1) = lim
x→0
1
�����:
√1√
1 + x + 1
=1
2
