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Best answer
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limit =1/2.

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limx->0(1-cosx)/xsinx       applying L'Hospital rule

=limx->0sinx/(sinx+xcosx)   

=limx->0((sinx/x)*x)/[((sinx/x)*x)+xcosx]

=limx->0x/(x+xcosx)

=limx->0 1/(1+cosx)    putting x=0

=1/1+1

=1/2

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