F07 Math 1205 -- Test 1 19 Sep 2007
NAME: SOLUTIONS ____ CRN:
You will be graded on how well and fully you support your work! Use only methods from class.
1
2
3
4
5
6
x
1
2
f
H
x
L
1. [24] a) Let
f
H
x
L
be defined by the graph given above. Mark each statement true or false.
f HxLis continuous
at x =1
TRUE
lim
xÆ3
f HxLexists
and is = + •
FALSE
lim
xÆ4+f HxL=1
FALSE
lim
xÆ2-f HxL=1
2
TRUE
f H1L+f H2L+f H3L
3=1
TRUE
lim
xÆ5
f HxLexists
TRUE
x=3 is a vertical
asymptote
TRUE
lim
xÆ2-f HxL≠lim
xÆ2+f HxL
TRUE
f HxLis discontinuous
at x =1, 2, 3, 4, 5
FALSE
x=5 is a removable
discontinuity of f
TRUE
lim
xÆ3
f HxLdoes not exist
TRUE
x=4 is a removable
discontinuity of f
FALSE
2. [9] Determine the value(s) of
a
for which
lim
x
Ø2
x
2-
a x
+
a
+1
x
-2
must exist. Evaluate the limit in this case(s).
Since the denominator vanishes at
x
=2
, the only chance that the limit has to exist is for the numerator to vanish at
x
=2
as well. Thus,
22-2
a
+
a
+1=0
, or after rearranging, a=5. The expression becomes:
x
2-5
x
+6
x
-2=H
x
-2L H
x
-3L
x
-2
, so the
limit exists for this value.
lim
x
Ø2
x
2-
a x
+
a
+1
x
-2
=
lim
x
Ø2
x
2-5
x
+6
x
-2=lim
x
Ø2
H
x
-2L H
x
-3L
x
-2=lim
x
Ø2H
x
-3L= -1
3. [9] Suppose that
H
x
-1L2§
f
H
x
L§cos p
x
2
for all
x
œ@0, 1D
. For what value(s) of c
must
lim
x
Ø
cf
H
x
L
exist? Determine the
corresponding limit value(s).
It is apparent from the setup that the Sandwich Theorem is called for. To see where the pinching occurs, let's do a
quick plot:
0.2 0.4 0.6 0.8 1.0
0.2
0.4
0.6
0.8
1.0
The blue curve is cos(p x/2) and the red curve is
H
x
-1L2
.
It's clear from the graph that sandwiching occurs only at x=0 and x=1. In fact, the hypothesis of the Sandwich Theorem
apply at these two points. Finally,
lim
x
Ø0
f
H
x
L=1
and
lim
x
Ø1
f
H
x
L=0
4. [21] Evaluate the following limits. Use only algebraic methods (i.e., no tables, no decimals). A bare answer is not
adequate; justify your work.
a)
lim
x
Ø-¶
x
3-
x
2
x
3+
x
2=lim
x
Ø-¶
1-1
x
2
2+1
x
=1
2
by the basic limit laws theorem and the fact that
lim
x
Ø-¶
1
x
=0
.
b)
lim
x
Ø-2+
x
2-3
x
+2
x
2H
x
+2L
=lim
x
Ø-2+
H
x
-2L I
x
-1M
x
2H
x
+2L
Since the denominator vanishes at the limit point without the benefit of the numera-
tor vanishing, the limit does not exist. We need only check whether its -¶ or +¶. At x a little greater than -2, x-2 is
negative, x-1 is negative and x+2 is positive as is
x
2
. Thus
lim
x
Ø-2+
x
2-3
x
+2
x
2H
x
+2L
=+¶.
c)
lim
x
Ø0
x
sinHp
x
L
1-cosH2
x
L
[Hint:
cosH2
x
L=1-2 sin2H
x
L
] Using the hint (sorry about that typo), we have:
lim
x
Ø0
x
sinHp
x
L
1-cosH2
x
L
=
lim
x
Ø0
x
sinHp
x
L
2 sin2H
x
L
=
lim
x
Ø0
x
sinHp
x
L
2 sin
x
sin
x
=
lim
x
Ø0
1
2
x
sin
x
sin px
sin
x
=lim
x
Ø0
p
2
x
sin
x
sin px
p
x
x
sin
x
As
x
Ø0
, each of the three factors
on the right tends to 1! (As derived in class.). Therefore,
lim
x
Ø0
x
sinHp
x
L
1-cosH2
x
L=p
2
.
