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lim xcosx-log(1+x)/x2x_0
lim (xcosx-log(1+x))/x2on putting x=0 we get 0/0 which is undefinedso by using L'Hospital rulelim {-xsinx+cosx-1/(1+x)}/2x again 0/0again using L'hospital rulelim {-xcosx-sinx-sinx+1/(1+x)2}/2=1/2Thanks & RegardsRinkoo GuptaAskIITians faculty
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