lim(x->0)[(x^2)*sin(1/x)]/sinx I use the equivalent infinitesimal, because when (x - > 0), SiNx ~ x, so sin (1 / x) ~ 1 / x, Then the original formula = LIM (x - > 0) (x ^ 2) * (1 / x) / SiNx = 1, right

lim(x->0)[(x^2)*sin(1/x)]/sinx I use the equivalent infinitesimal, because when (x - > 0), SiNx ~ x, so sin (1 / x) ~ 1 / x, Then the original formula = LIM (x - > 0) (x ^ 2) * (1 / x) / SiNx = 1, right

incorrect.
"I use the equivalent infinitesimal, because when (x - > 0), SiNx ~ x, so sin (1 / x) ~ 1 / X", here (x - > 0), SiNx ~ x, correct, but when
1 / x, approaching infinity, so it doesn't hold
The correct answer should be
lim (x/sinx)*(xsin(1/x))
=lim 1*(xsin1/x)
=0
The limit of infinitesimal x bounded function (sin1 / X bounded) is 0