Aditya Gupta
Last Activity: 4 Years ago
statement 2 is clearly true as it is a std property/theorem of limits.
in statement 1, if we let h(x)= [x] and g(x)= sinx/x, then [sinx/x]= h(g(x)).
hence lim x tends to 0 [sinx/x] and [lim x tends to 0 (sinx/x)] would have been equal if h(x) were continuous at x= lim x tends to 0 g(x)= x= lim x tends to 0 sinx/x= 1. but obviously h(x)= [x] is discontinuous at x= 1.
in fact using sinx/x= sin|x|/|x| and sin|x| is less than |x| for x not equal to 0 (this can be observed from graph), we can say that lim x tends to 0 [sinx/x]= 0 and [lim x tends to 0 (sinx/x)]= [1]= 1. so they arent equal.
hence option A.
KINDLY APPROVE ;))