Limits and Continuity 71- Rewrite limx→b
x^3 −b^3
x^6 −b^6
aslimx→b
x^3 −b^3
(x^3 −b^3 )(x^3 +b^3 )
=limx→b1
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 obtainlimx→ 02 −
√
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 asxlim→−∞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)
=eandxlim→ 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→ 0sin 3x
x
sin 4x
x. Now rewrite
the limit as limx→ 03
sin 3x
3 x
4
sin 4x
4 x=
3
4
limx→ 0sin 3x
3 x
sin 4x
4 x.
Asxapproaches 0, so do 3xand 4x.
Thus, you have3
4
3 limx→ 0sin 3x
3 x4 limx→ 0sin 4x
4 x