1400 Quiz 2
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Pennsylvania State University *
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111
Subject
Mathematics
Date
Feb 20, 2024
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4
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Name:
Math 1400
Quiz #2
Penn ID:
8:30am
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Math 1400
Quiz #2
Penn ID:
8:30am
1. Evaluate the following limit using any methods that you know:
lim
x
!
0
ln(1 + sin
2
(
x
))
3 sinh
2
(
x
)
Solution: Note that
3 sinh
2
(
x
) = 3
✓
x
+
x
3
3!
+
x
5
5!
+
. . .
◆
2
= 3
x
2
+ H.O.T.
Recall the Taylor series
ln(1 +
x
) =
x
-
x
2
2
+
x
3
3
-
. . . ,
which converges when
|
x
|
<
1. Using sin
2
(
x
) in place of
x
, which is allowed because sin
2
(
x
)
<
1 (as it actually converges to zero) when
x
!
0 and then plugging in the Taylor series for
sin(
x
), we have:
ln(1 + sin
2
(
x
)) = sin
2
(
x
)
-
(
sin
2
(
x
)
)
2
2
+
(
sin
2
(
x
)
)
3
3
-
. . .
=
x
2
+ H.O.T.
Thus:
lim
x
!
0
ln(1 + sin
2
(
x
))
3 sinh
2
(
x
)
= lim
x
!
0
x
2
(1 + H.O.T.)
3
x
2
(1 + H.O.T.)
=
1
3
·
lim
x
!
0
1 + H.O.T.
1 + H.O.T.
=
1
3
·
1 = 1
.
Math 1400
Quiz #2
Penn ID:
8:30am
2. Consider the function
f
(
x
) = (1 + 2 sin(
x
))
1
2
.
Find a degree-three polynomial that can be used as an approximation to
f
(
x
) when
x
is close
to zero.
Solution: The degree-three Taylor polynomial would be a good approximation to
f
(
x
) when
x
is close to zero. We compute this polynomial by first using the Binomial series (thinking of
2 sin(
x
) as the variable), and then using the Taylor series for sin(
x
).
We perform these calculations as follows:
f
(
x
) = (1 + 2 sin(
x
))
1
2
= 1 +
1
2
·
(2 sin(
x
)) +
1
2
(
1
2
-
1
)
2!
(2 sin(
x
))
2
+
1
2
(
1
2
-
1
) (
1
2
-
2
)
3!
(2 sin(
x
))
3
+
. . .
= 1 + sin(
x
)
-
sin
2
(
x
)
2
+
sin
3
(
x
)
2
-
. . .
= 1 +
✓
x
-
x
3
3!
+ H.O.T.
◆
-
1
2
✓
x
-
x
3
3!
+ H.O.T.
◆
2
+
1
2
✓
x
-
x
3
3!
+ H.O.T.
◆
3
-
. . .
Since we only care about terms upto the third degree, we collect all those terms to find our
desired approximation:
f
(
x
)
⇡
1 +
x
-
x
3
6
-
1
2
·
x
2
+
1
2
·
x
3
=
x
3
3
-
x
2
2
+
x
+ 1
.
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Math 1400
Quiz #2
Penn ID:
8:30am