MA 3972-MA-Book May 7, 2018 9:52
106 STEP 4. Review the Knowledge You Need to Score High
Thus, you have
3
4
3 limx→ 0
sin 3x
3 x
4 limx→ 0
sin 4x
4 x
=
3 ( 1 )
4(1)
=
3
4
.
- Ast→ 3 +,(t−3)>0, and thus
(t− 3 )=
√
(t− 3 )^2. Rewrite the limit as
limt→ 3 +
√
(t− 3 )(t+ 3 )
√
(t− 3 )^2
=tlim→ 3 +
√
(t+ 3 )
√
(t− 3 )
.
The limit of the numerator is
√
6, and the
denominator is approaching 0 through
positive values. Thus, limt→ 3 +
√
t^2 − 9
t− 3
=∞.
- The graph off indicates that:
I. limx→ 4 − f(x)=5 is true.
II. lim
x→ 4
f(x)=2 is false.
(The limx→ 4 f(x)= 5 .)
III. “x=4 is not in the domain of f”is
false sincef(4)=2.
Part B Calculators are allowed.
- Examining the graph in your calculator,
you notice that the function approaches
thex-axis asx→∞or asx→−∞.
Thus, the liney=0 (thex-axis) is a
horizontal asymptote. Asxapproaches 1
from either side, the function increases or
decreases without bound. Similarly, asx
approaches−2 from either side, the
function increases or decreases without
bound. Therefore,x=1 andx=−2 are
vertical asymptotes. (See Figure 6.7-1.)
[–6, 5] by [–3, 3]
Figure 6.7-1
- Asx→ 5 +, the limit of the numerator
(5+[5]) is 10 and asx→ 5 +, the
denominator approaches 0 through
negative values. Thus, the lim
x→ 5 +
5 +[x]
5 −x
=−∞.
- Since f(x) is a rational function, it is
continuous everywhere except at values
where the denominator is 0. Factoring and
setting the denominator equal to 0, you
have (x+6) (x−2)=0. Thus, the
function is discontinuous atx=−6 and
x=2. Verify your result with a calculator.
(See Figure 6.7-2.)
[–8, 8] by [–4, 4]
Figure 6.7-2
- In order forg(x) to be continuous at
x=3, it must satisfy the three conditions
of continuity:
(1)g(3)= 32 + 5 =14,
(2) limx→ 3 +(x^2 +5)=14, and
(3) limx→ 3 −(2x−k)= 6 −k, and the two
one-sided limits must be equal in order for
limx→ 3 g(x) to exist. Therefore, 6−k= 14
andk=−8.
Now,g(3)=limx→ 3 g(x) and condition 3 is
satisfied.
- Checking with the three conditions of
continuity:
(1)f(2)=12,
(2) limx→ 2
x^2 + 5 x− 14
x− 2
=
lim
x→ 2
(x+7)(x−2)
x− 2
=lim
x→ 2
(x+7)=9, and
(3)f(2)=limx→ 2 (x+7). Therefore,f(x)
is discontinuous atx=2.
