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Byju's Answer
Standard XII
Mathematics
Algebra of Limits
x→ 0lim sin x...
Question
l
i
m
x
→
0
(
sin
x
x
)
sin
x
x
−
sin
x
=
2
a
e
Find
a
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Solution
l
i
m
x
→
0
(
sin
x
x
)
sin
x
x
−
sin
x
=
l
i
m
x
→
0
[
1
+
sin
x
−
x
x
]
sin
x
−
(
sin
x
−
x
)
=
l
i
m
x
→
0
[
1
+
sin
x
−
x
x
]
x
sin
x
−
x
.
−
sin
x
x
=
e
l
i
m
x
→
0
−
sin
x
x
=
1
e
∴
Required value of
a
is 2.
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0
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Q.
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sin
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