Question

f(x) = { x^2sin(1/x), if x =/= 0
0, if x=0}

a. find lim(x->0)f(x) and show that f(x) is continuous at x=0.


b, find f'(0) using the definition of the derivative at x=0:

f'(x)=lim(x->0) (f(x)-f(0)/x)

c. Show that lim(x->0)f'(x) does not exist. In particular, this means lim(x->0)f'(x) =/= f'(0), hence f'(x) is not continuous at x=0.

Answer 1

since |sinx| <= 1, x^2 sin(1/x) < x^2. So, since x^2 -> 0, so does x^2 sin(1/x)

x^2 sin(1/x) does not exist when x=0, but since f(0)=0, f is continuous.

[(0+h)^2 sin(1/(0+h)) - 0^2*sin(1/0)]/h
= [h^2 sin(1/h) - 0]/h
= h^2 ((1/h) - 1/3! (1/h^3) + ...)/h
= (h - h/3! + ...)/h
= 1 - 1/3! + ...
-> 5/6

using l'Hospital's rule

f'(x) = 2x sin(1/x) - cos(1/x)
2x sin(1/x) -> 0 but
cos(1/x) is undefined, somewhere between -1 and 1.

so, I guess 5/6 is as good a value as any.