handa (nh5757) – Limits & continuity – bormashenko – (54880)
1
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printout
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have
16
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
Below is the graph of a function
f
.
2
4
6

2

4

6
2
4
6
8

2

4
Use the graph to determine all the values of
x
on (

6
,
6) at which
f
fails to be continuous.
1.
x
=

3
2.
no values of
x
3.
x
= 4
4.
x
=

3
,
4
correct
5.
none of the other answers
Explanation:
Since
f
(
x
) is defined for all values of
x
on (

6
,
6), the only values of
x
in (

6
,
6)
at which the function
f
is discontinuous are
those for which
lim
x
→
x
0
f
(
x
) =
f
(
x
0
)
or
lim
x
→
x
0

f
(
x
) =
lim
x
→
x
0
+
f
(
x
)
.
The only possible candidates here are
x
0
=

3 and
x
0
= 4. But at
x
0
=

3
f
(

3) = 8 =
lim
x
→ 
3
f
(
x
) = 4
,
while at
x
0
= 4
lim
x
→
4

f
(
x
) = 6 =
lim
x
→
4+
f
(
x
) = 4
.
Consequently, on (

6
,
6) the function
f
fails
to be continuous only at
at
x
=

3
,
4
.
002
10.0 points
Use continuity to evaluate
lim
x
→
3
π
sin (
x
+ 4 sin
x
)
.
1.
limit =
∞
2.
limit = 0
correct
3.
limit = 3
π
4.
limit = 1
5.
limit =

1
Explanation:
Because both
x
are sin
x
is continuous on
(
∞
,
∞
), the sum
x
+ 4 sin
x
also is contin
uous everywhere on (
∞
,
∞
). But then the
composition
f
(
x
) = sin(
x
+ 4 sin
x
)
too is continuous everywhere on (
∞
,
∞
).
Now by definition,
lim
x
→
c
f
(
x
) =
f
(
c
)
whenever
f
is continuous at
x
=
c
.
For the
given function
f
, therefore,
lim
x
→
3
π
f
(
x
) = sin(3
π
+ 4 sin(3
π
))
.
Consequently,
limit = 0
.
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handa (nh5757) – Limits & continuity – bormashenko – (54880)
2
003
10.0 points
Find all the points at which the function
f
(
x
) =
x
2

2
x

8
is not continuous.
1.
x
= 4
2.
x
=

2
3.
continuous everywhere
correct
4.
none of these
5.
x
= 4
,

2
Explanation:
Since
f
(
x
) =
x
2

2
x

8
is a polynomial function, it is
continuous everywhere
,
including the points
x
= 4
,

2 at which
f
(4) =
f
(

2) = 0
.
004
10.0 points
Find all values of
x
at which the function
f
defined by
f
(
x
) =
x

6
x
2

x

30
is continuous, expressing your answer in in
terval notation.
1.
(
∞
,
6)
∪
(6
,
∞
)
2.
(
∞
,

5)
∪
(

5
,

6)
∪
(

6
,
∞
)
3.
(
∞
,

6)
∪
(

6
,
5)
∪
(5
,
∞
)
4.
(
∞
,

5)
∪
(

5
,
∞
)
5.
(
∞
,

5)
∪
(

5
,
6)
∪
(6
,
∞
)
correct
Explanation:
After
factorization the
denominator be
comes
x
2

x

30 = (
x

6)(
x
+ 5)
,
so
f
can be rewritten as
f
(
x
) =
x

6
(
x

6)(
x
+ 5)
=
1
(
x
+ 5)
whenever
x
= 6. At
x
= 6 both the numer
ator and denominator will be zero; thus
f
will not be defined, hence not continuous, at
x
= 6.
Elsewhere
f
is a ratio of polynomial
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 Spring '10
 Gualdini
 Topology, Continuity, Limits, Limit, Continuous function, lim g

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