The problem is that you can't be sure that x^2 sin(1/x) goes to 0, since sin(1/x) does not have a limit at 0... So you can't use L'Hospital.
I'm having trouble calculating the following limit:
limit as x goes to zero of ( x^2 * sin(1/x) ) / sinx
I can see that it's an indeterminate form of 0/0 so it seems like classic L'Hospital,
but when I take the limit of the derivatives it comes out:
(2x * sin(1/x) - cos(1/x) ) / cosx
and cos(1/x) has no limit at zero, so I don't know what to do.