is Sin^2 = 2sin?

Question

the question is limx^3sin^2(1/x)
when x->0

Answer 1

No. sin² x is (sin x)², whereas 2 sin x is sin x + sin x. As for your limit:

The sin function, and therefore sin² (1/x), will never leave the range [-1, 1], no matter what 1/x is. So x³ sin² (1/x) is always strictly between -x³ and x³, both of which approach 0 as x→0, and thus by the squeeze theorem [x→0]lim x³ sin² (1/x) = 0.

Answer 2

limx>0x^3sin^2(1/x)
=limx>0 x[sin^2(1/x)/1/x^2]
=limx>0x*limx>0[sin^2(1/x)/(1/x^2)]=0

Answer 3

no they arnt
2sinx woudld be like 2*sinx instead of sinx*sinx