Question #047be

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Answer 1 · 2016-01-05T20:59:48

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 ;)