lim x tends to 0 (1+cos 4x)(1-cos3x)/x^2 =?
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Answer:
the answer
lim
x
→
0
16
cos
(
4
x
)
−
9
c
o
(
3
x
)
2
=
16
−
9
2
=
7
2
Explanation:
show below
lim
x
→
0
cos
(
3
x
)
−
cos
(
4
x
)
x
2
Direct compensation product equal
0
0
so we will use l'hospital rule
lim
x
→
a
f
(
x
)
g
(
x
)
=
0
0
lim
x
→
a
f
'
(
x
)
g
'
(
x
)
f
(
x
)
=
cos
(
3
x
)
−
cos
(
4
x
)
f
'
(
x
)
=
4
sin
(
4
x
)
−
3
sin
(
3
x
)
g
(
x
)
=
x
2
g
'
(
x
)
=
2
x
lim
x
→
0
4
sin
(
4
x
)
−
3
sin
(
3
x
)
2
x
since the Direct compensation product again equal
0
0
so we will derive it again
f
'
'
(
x
)
=
16
cos
(
4
x
)
−
9
c
o
(
3
x
)
g
'
'
(
x
)
=
2
lim
x
→
0
16
cos
(
4
x
)
−
9
c
o
(
3
x
)
2
=
16
−
9
2
=
7
2
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