What is the limit of #lim_((x,y)->(0,0))x^3/(x^2+y^2)#?

In order for this limit to exist, the fraction #x^2/(x^2+y^2)#must approach the same value #L#

regardless of the path along which we approach#(0,0)#
Consider approaching#(0,0)# along the x-axis That means fixing
#y=0# and finding the limit #color(red)[lim_(x->0) x^2/(x^2+y^2)]#

#lim_{x->0," "y=0} x^2/(x^2+y^2)=lim_(x->0)x^2/(x^2+0)=0/0#

since the direct compensation product equal #0/0# we will use
L'Hôpital's rule

#lim_[(x)rarr(a)][f'(x)]/[g'(x)]#

#lim_(x->0)x^2/(x^2+0)=lim_(x->0)(2x)/(2x)=0/0#

since the direct compensation product also equal #0/0# we will use
L'Hôpital's rule again

#lim_(x->0)(2x)/(2x)=lim_(x->0)2/2=1#

Now, consider approaching,(0,0) along the y-axis This means fixing x=0 and finding the limit #color(red)[lim_(y->0) x^2/(x^2+y^2)]#

#lim_{y->0," "x=0} x^2/(x^2+y^2)=lim_(y->0)0/(0+y^2)=0/0#

#lim_(y->0)0/(0+y^2)=lim_(y->0)0/(2y)=lim_(y->0)0/2=0#

Approaching the origin along these two different paths leads to different limits

#color(blue)[lim_{x->0," "y=0} x^2/(x^2+y^2)!=lim_{y->0," "x=0} x^2/(x^2+y^2)]#

Thus, the original limit does not exist.