72 STEP 4. Review the Knowledge You Need to Score High
- Ast→ 3 +,(t−3)>0, and thus
(t− 3 )=
√
(t− 3 )^2. Rewrite the limit astlim→ 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.xlim→ 4 −f(x)=5 is true.
II.xlim→ 4 f(x)=2 is false.
(The limx→ 4 f(x)= 5 .)
III. “x=4 is not in the domain off”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 5.7-1.)
[–6,5] by [–3,3]
Figure 5.7-1- Asx→ 5 +, the limit of the numerator
(5+[5]) is 10 and asx→ 5 +, the
denominator approaches 0 through
negative values. Thus, the
xlim→ 5 +5 +[x]
5 −x=−∞.
- Sincef(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
the function is discontinuous atx=− 6
andx=2. Verify your result with a
calculator. (See Figure 5.7-2.)
[–8,8] by [–4,4]
Figure 5.7-2- 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
=
xlim→ 2(x+7)(x−2)
x− 2
=xlim→ 2 (x+7)=9, and
(3) f(2)/=limx→ 2 (x+7). Therefore,f(x)is
discontinuous atx=2.- The graph indicates that (a) f(3)=4,
(b) limx→ 3 +f(x)=0, (c) limx→ 3 − f(x)=0,
(d) limx→ 3 f(x)=0, and (e) therefore,f(x)
is not continuous atx=3 since
f(3)/=limx→ 3 f(x).