at x = c not= 0, f(x) is continuous and hence differentiable.

so f(c) = (c^2)sin(1/c)

=> f ' (c) = 2csin(1/c) + (-c^-2)(c^2)cos(1/c)

=> f ' (c) = 2csin(1/c) - cos(1/c)

we must show that f is continuous at x = 0B) show that f is differentiable at x = 0 and show that f(0)=0

note that |(x^2)sin(1/x)|<or= x^2 for all x

so we have: -x^2 <or= (x^2)sin(1/x) <or= x^2

=> lim{x-->0+} -x^2 <or= lim{x-->0+}(x^2)sin(1/x) <or= lim{x-->0+}x^2

=> 0 <or= lim{x-->0+}(x^2)sin(1/x) <or= 0

so by the squeeze theorem:

lim{x-->0+}(x^2)sin(1/x) = 0

similarly, we can show, lim{x-->0-}(x^2)sin(1/x) = 0

thus f is continuos at x = 0, so by the definition of continuity we have f(0) = lim{x-->0}(x^2)sin(1/x) = 0

this seems trivial to me, but i may be wrong. at x = 0, 1/x is undefined, so f ' (x) = 2xsin(1/x) - cos(1/x) is undefined and is not continuous. you can take the left and right hand limits to see this, show that they are different or somethingC) show that f' is not continuous at x = 0