The given function is
f(x,y)=(y+x)(y−x)×(1+y2)(1+x2)
The repeated limit of f(x,y) are following
limx→0limy→0f(x,y)=limx→0(limy→0f(x,y))
=limx→0(limy→0(y+x)(y−x)×(1+y2)(1+x2))
=limx→0(−1)(1+x2)
=−1.
Again, limy→0limx→0f(x,y)=limy→0(limx→0f(x,y))
=limy→0(limx→0(y+x)(y−x)×(1+y2)(1+x2))
=limy→01+y21
=1
Hence ,the repeated limits exist and unequal.
Now we have to find the simultaneous limit of f(x,y)
lim(x,y)→(0,0)f(x,y)=lim(x,y)→(0,0)(y+x)(y−x)×(1+y2)(1+x2)
Let (x,y)→(0,0) along the line y=mx ,m∈R then
lim(x,y)→(0,0)f(x,y)=limx→0(mx+x)(mx−x)×(1+m2x2)(1+x2)
=(1+m)(1−m)×limx→0(1+m2x2)(1+x2)=(1+m)(1−m)Which is not unique as it takes different values for different values of m .
Hence the simultaneous limit does not exist.
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