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KEAM 2015 Math Paper
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© examsnet.com
Question : 108
Total: 120
lim
x
→
0
(
10
9
s
i
n
9
x
s
i
n
10
x
)
(
8
7
s
i
n
7
x
s
i
n
8
x
)
(
6
5
s
i
n
5
x
s
i
n
6
x
)
(
4
3
s
i
n
3
x
s
i
n
4
x
)
(
s
i
n
x
s
i
n
2
x
)
=
63
256
1
6
6
5
1
2
256
63
Validate
Solution:
lim
x
→
0
(
10
9
sin
9
x
sin
10
x
)
(
8
7
sin
7
x
sin
8
x
)
(
6
5
sin
5
x
sin
6
x
)
(
4
3
sin
3
x
sin
4
x
)
(
sin
x
sin
2
x
)
=
lim
x
→
0
(
sin
9
x
9
x
⋅
1
sin
10
x
10
x
)
(
sin
7
x
7
x
⋅
1
sin
8
x
8
x
)
(
sin
5
x
5
x
⋅
1
sin
6
x
6
x
)
(
sin
3
x
3
x
⋅
1
sin
4
x
4
x
)
(
sin
x
x
1
sin
2
x
2
x
⋅
1
2
)
=
1
2
[
∵
lim
f
(
x
)
→
0
sin
f
(
x
)
f
(
x
)
=
1
]
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