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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
D.0 use expansion of sinx
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
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
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
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
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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