Question

Question #047be

Answer 1

Yep i can

`u = xy`

`f(u) = u*ln(u)`

`f(u) = ln(u)/(1/u)`

`lim_(u -> 0) f(u)` ?

Hôpital rule's

`(1/u)/(-1/u^2) = -u`

`lim_(u -> 0) (-u) = 0`

so `lim_((x,y) -> (0,0)) f(x,y) = 0`

I let you do the second ;)