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Byju's Answer
Standard XI
Mathematics
Expansions to Remove Indeterminate Form
x → 0limesin ...
Question
lim
x
→
0
e
sin
x
−
sin
x
−
1
x
2
is equal to
A
1
3
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B
1
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C
e
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D
1
2
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Solution
The correct option is
D
1
2
lim
x
→
0
e
sin
x
−
sin
x
−
1
x
2
=
lim
x
→
0
(
1
+
sin
x
1
!
+
sin
2
x
2
!
+
sin
3
x
3
!
+
.
.
.
)
−
sin
x
−
1
x
2
=
lim
x
→
0
sin
2
x
x
2
(
1
2
!
+
sin
x
3
!
+
.
.
.
)
=
1
2
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0
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