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Question

The value of limx02cosx1sin1(1t2)dt2xsin2x is

A
0
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B
does not exist
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C
1
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D
12
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Solution

The correct option is B does not exist
limx02cosx1sin1(1t2)dt2xsin2x
On applying limx0, we get 00 form.
Therefore, use L'Hospitals' rule,
limx02(sin11cos2 x)(sin x)22 cos 2x
=limx0(sin1|sinx|)sin x2 sin2 x

=limx0sin1|sinx|2 sin x
L.H.L =limx0(x)2 sin x=12
R. H. L =limx0+(x)2 sin x=12.
L. H. L R.H.L
limit does not exist.

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