(a)
lim
sin x
is equal to
0X (1 + cos x)
2.
1
Answers
Step-by-step explanation:
The Taylor Expansions for sine and cosine are:
sin
(
u
)
=
u
−
u
3
3
!
+
O
(
u
5
)
and
cos
(
u
)
=
1
−
u
2
2
!
+
O
(
u
4
)
Plugging these into the limit yields:
lim
x
→
0
x
2
−
x
6
3
!
+
O
(
x
10
)
1
−
(
1
−
x
2
2
!
+
O
(
x
4
)
)
=
lim
x
→
0
x
2
−
x
6
3
!
+
O
(
x
10
)
x
2
2
!
+
O
(
x
4
)
Divide top and bottom by
x
2
:
=
lim
x
→
0
1
+
O
(
x
4
)
1
2
!
+
O
(
x
2
)
=
2
Answer link
Answer:
Use the fundamental trigonometric limits and algebra.
Explanation:
The fundamental trigonometric limits are:
lim
x
→
0
cos
x
−
1
x
=
0
and
lim
x
→
0
sin
x
x
=
1
If the problem is
cos
x
−
1
sin
(
x
2
)
, then use
cos
x
−
1
sin
(
x
2
)
=
cos
x
−
1
x
2
⋅
x
2
sin
(
x
2
)
So the limit we seek is equal to
lim
x
→
0
cos
x
−
1
x
2
Looking for a trick that might help, we it may eventually occur to us to try mulltiplying by
cos
x
+
1
cos
x
+
1
to replace the subtraction that goes to
0
with an addition that goes to
2
.
cos
x
−
1
x
2
cos
x
+
1
cos
x
+
1
=
cos
2
x
−
1
x
2
(
cos
x
+
1
)
=
−
sin
2
x
x
2
(
cos
x
+
1
)
=
−
sin
x
x
⋅
sin
x
x
⋅
1
cos
x
+
1
The limit as
x
→
0
is
−
(
1
)
⋅
(
1
)
⋅
1
1
+
1
=
−
1
2
Bonus
If the problem is intended to be
cos
x
−
1
(
sin
x
)
2
, then use the same trick to get:
cos
x
−
1
(
sin
x
)
2
=
cos
x
−
1
(
sin
x
)
2
cos
x
+
1
cos
x
+
1
=
−
(
sin
x
)
2
(
sin
x
)
2
(
cos
x
+
1
)
=
−
1
cos
x
+
1
.
Step-by-step explanation: