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limit x tends to 0 ((x - sin x)/x)sin(1/x) A.1 B.does not exist C.-1 D.0

limit x tends to 0 ((x - sin  x)/x)sin(1/x)


A.1


B.does not exist


C.-1


D.0

Grade:12th Pass

4 Answers

Harshit Mathur
20 Points
10 years ago

D.0  use expansion of sinx

Neha Gaur
6 Points
10 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)]

                         [0][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)]

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

RHL=LHL=0

Aditya Verma
33 Points
10 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

Amlan Kumar
39 Points
5 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
 
 
 
 
 

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