Limits and Continuity 71
- Rewrite limx→b
x^3 −b^3
x^6 −b^6
as
limx→b
x^3 −b^3
(x^3 −b^3 )(x^3 +b^3 )
=limx→b
1
x^3 +b^3
.
Substitutex=band obtain
1
b^3 +b^3
=
1
2 b^3
.
- Substitutingx=0 into the expression
2 −
√
4 −x
x
leads to 0/0, which is an
indeterminate form. Thus, multiply both
the numerator and denominator by the
conjugate
(
2 +
√
4 −x
)
and obtain
limx→ 0
2 −
√
4 −x
x
(
2 +
√
4 −x
2 +
√
4 −x
)
=limx→ 0
4 −(4−x)
x
(
2 +
√
4 −x
)
=limx→ 0
x
x
(
2 +
√
4 −x
)
=limx→ 0
(^1
2 +
√
4 −x
)
=
1
(
2 +
√
4 −( 0 )
)=
1
4
.
- Since the degree of the polynomial in the
numerator is the same as the degree of
the polynomial in the denominator,
xlim→∞
5 − 6 x
2 x+ 11
=−
6
2
=−3.
- Since the degree of the polynomial in the
numerator is 2 and the degree of the
polynomial in the denominator is 3,
xlim→−∞
x^2 + 2 x− 3
x^3 + 2 x^2
=0.
- The degree of the monomial in the
numerator is 2 and the degree of the
binomial in the denominator is 1. Thus,
xlim→∞
3 x^2
5 x+ 8
=∞.
- Divide every term in both the numerator
and denominator by the highest power
ofx. In this case, it isx. Thus, you have
xlim→−∞
3 x
√x
x^2 − 4
x
.Asx→−∞,x=−
√
x^2.
Since the denominator involves a radical,
rewrite the expression as
xlim→−∞
3 x
√x
x^2 − 4
−
√
x^2
=xlim→−∞
3
−
√
1 −
4
x^2
=
3
−
√
1 − 0
=− 3.
- xlim→ 1 +f(x)=xlim→ 1 +
(
x^2 ex
)
=eand
xlim→ 1 −f(x)=xlim→ 1 −(ex)=e. Thus,
xlim→ 1 f(x)=e.
- xlim→∞ex=∞and limx→∞
(
1 −x^3
)
=∞.
However, asx→∞, the rate of increase
ofexis much greater than the rate of
decrease of (1−x^3 ). Thus,
xlim→∞
ex
1 −x^3
=−∞.
- Divide both numerator and denominator
byxand obtain limx→ 0
sin 3x
x
sin 4x
x
. Now rewrite
the limit as limx→ 0
3
sin 3x
3 x
4
sin 4x
4 x
=
3
4
limx→ 0
sin 3x
3 x
sin 4x
4 x
.
Asxapproaches 0, so do 3xand 4x.
Thus, you have
3
4
3 limx→ 0
sin 3x
3 x
4 limx→ 0
sin 4x
4 x