Q   72 .   T h e   v a l u e   o f   lim x 0 +   1 cos   x   cos - 1 t   d t 2 x - sin   2 x   i s   e q u a l   t o : A   0                                                     B   - 1 C   2 3                                                 D   - 1 4

Dear Student,
Please find below the solution to the asked query:

L=limx0+ 1cosxcos-1tdt2x-sin2xAs cos0=1 , hence limit will become 1 to 1 which means integral becomes0 when x=0 and denominator is also 0 at x=0, hence we have 00 formi.e. L-Hopital rule can be applied. Use Lebnitz theorem in numerator to differentiateL= limx0+ cos-1cosxddxcosx-02-2cos2x=-12 limx0+ xsinx1-cos2x=-12 limx0+ xsinx1-1+2sin2x=-14limx0+ xsinx=-14×1=-14

Hope this information will clear your doubts about this topic.

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