Answer to Question #108248 in Calculus for TUHIN SUBHRA DAS

Question #108248
find the two repeated limits of the function f(x,y)=(y-x/y+x)(1+x^2/1+y^2) at (0,0).Does the simultaneous limit of f exist as (x,y) tends to (0,0)?
1
Expert's answer
2020-04-07T12:38:09-0400

The given function is


f(x,y)=(yx)(y+x)×(1+x2)(1+y2)f(x,y)=\frac{(y-x)}{(y+x)} ×\frac{(1+x^2)}{(1+y^2)}


The repeated limit of f(x,y)f(x,y) are following

limx0limy0f(x,y)=limx0(limy0f(x,y))\lim_{x\to 0} \lim_{y\to 0} f(x,y)=\lim_{x\to 0}(\lim_{y\to 0} f(x,y))

=limx0(limy0(yx)(y+x)×(1+x2)(1+y2))=\lim_{x\to 0}(\lim_{y\to 0} \frac{(y-x)}{(y+x)}×\frac{(1+x^2)}{(1+y^2)})

=limx0(1)(1+x2)=\lim_{x\to 0} (-1) (1+x^2)

=1.=-1.

Again, limy0limx0f(x,y)=limy0(limx0f(x,y))lim_{y\to 0}lim_{x\to 0}f(x,y)=\lim_{y\to 0}(\lim_{x\to 0} f(x,y))

=limy0(limx0(yx)(y+x)×(1+x2)(1+y2))=\lim_{y\to0}(\lim_{x\to0} \frac{(y-x)}{(y+x)}×\frac{(1+x^2)}{(1+y^2)} )

=limy011+y2=\lim_{y\to0}\frac{1}{1+y^2}

=1=1

Hence ,the repeated limits exist and unequal.

Now we have to find the simultaneous limit of f(x,y)f(x,y)


lim(x,y)(0,0)f(x,y)=lim(x,y)(0,0)(yx)(y+x)×(1+x2)(1+y2)\lim_{(x,y)\to (0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{(y-x)}{(y+x)}×\frac{(1+x^2)}{(1+y^2)}


Let (x,y)(0,0)(x,y)\rightarrow(0,0) along the line y=mxy=mx ,mRm\in\R then


lim(x,y)(0,0)f(x,y)=limx0(mxx)(mx+x)×(1+x2)(1+m2x2)\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{x\to0} \frac{(mx-x)}{(mx+x)} ×\frac{(1+x^2)}{(1+m^2x^2)}




=(1m)(1+m)×limx0(1+x2)(1+m2x2)=\frac{(1-m)}{(1+m)}×\lim_{x\to0} \frac{(1+x^2)}{(1+m^2x^2)}=(1m)(1+m)=\frac{(1-m)}{(1+m)}

Which is not unique as it takes different values for different values of mm .

Hence the simultaneous limit does not exist.




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