162
Chapter 2 | Limits
To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.
⎛ ⎝ 2 x 2 −3 x +1 ⎞ ⎠
lim x →2
2 x 2 −3 x +1 x 3 +4
Apply the quotient law, making sure that. (2) 3 +4≠0
lim x →2
=
⎛ ⎝ x 3 +4
⎞ ⎠
lim x →2
x 2 −3· lim x →2
2· lim
x + lim x →2
1
x →2
=
Apply the sum law and constant multiple law.
x 3 + lim x →2
lim x →2
4
⎛ ⎝ lim
x ⎞ ⎠
2
2·
−3· lim x →2
x + lim x →2
1
x →2
=
Apply the power law.
⎛ ⎝ lim
x ⎞ ⎠
3
+ lim
4
x →2
x →2
= 2(4) − 3(2) + 1 (2) 3 +4
= 1 4 .
Apply the basic limit laws and simplify.
2.11
(2 x −1) x +4. In each step, indicate the limit law applied.
Use the limit laws to evaluate lim x →6
Limits of Polynomial and Rational Functions By now you have probably noticed that, in each of the previous examples, it has been the case that lim x → a f ( x ) = f ( a ). This is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a for which the rational function is defined.
Theorem 2.6: Limits of Polynomial and Rational Functions Let p ( x ) and q ( x ) be polynomial functions. Let a be a real number. Then, lim x → a p ( x ) = p ( a ) lim x → a p ( x ) q ( x ) = p ( a ) q ( a ) when q ( a ) ≠0.
n −1 + ⋯ + c
To see that this theorem holds, consider the polynomial p ( x ) = c n x n + c n −1 x
1 x + c 0 . By applying the
sum, constant multiple, and power laws, we end up with lim x → a p ( x ) = lim x → a ⎛ ⎝ c n x n + c n −1 x
⎞ ⎠
n −1 + ⋯ + c
1 x + c 0
n −1
n
⎛ ⎝ lim x → a
x ⎞ ⎠
⎛ ⎝ lim x → a
x ⎞ ⎠
⎛ ⎝ lim x → a
x ⎞
= c n
+ c n −1
+ ⋯ + c 1
⎠ + lim x → a
c 0
n −1 + ⋯ + c
n + c
= c n a
n −1 a
1 a + c 0
= p ( a ). It now follows from the quotient law that if p ( x ) and q ( x ) are polynomials for which q ( a ) ≠0, then lim x → a p ( x ) q ( x ) = p ( a ) q ( a ) . Example 2.16 applies this result.
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