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# limit x tends to 0 ((x - sin  x)/x)sin(1/x)A.1B.does not existC.-1D.0 9 years ago

lim x-->0 [(x-sinx)/x][sin(1/x)]

LHL->  lim h-->0 [(-h+sin h)/(-h)][sin(1/(-h)]

[((sin h)- h)/h][sin(1/h)]

[((sinh)/h)-1][sin(1/h)]

[sin(1/h)]=0                       {lim h-->0 (sin h)/h  =1}

RHL->  lim h-->0  [(h-sin h)/h][sin(1/h)]

[1- (sin h)/h][sin(1/h)]

[sin(1/h)]=0                          {"   "   "    "    "     "   "}

RHL=LHL=0

8 years ago

D.0

by using algebra of limits

=lim x tends to 0 (x/x - sinx/x) lim x tends to 0 sin(1/x).x/x   (multiply and divide by x)

=(1-1).0

=0

4 years ago
the concept is that the sin(infinity) has a finite value between -1 and 1 so cannot be treated as indeterminate form. so simplyfying and putting the limit we have 0*sin(infinity) which is 0