Answer to Question #108248 in Calculus for TUHIN SUBHRA DAS

Question

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)?

Answer 1

The given function is

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

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

$\lim_{x\to 0} \lim_{y\to 0} f(x,y)=\lim_{x\to 0}(\lim_{y\to 0} f(x,y))$

$=\lim_{x\to 0}(\lim_{y\to 0} \frac{(y-x)}{(y+x)}×\frac{(1+x^2)}{(1+y^2)})$

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

$=-1.$

Again, $lim_{y\to 0}lim_{x\to 0}f(x,y)=\lim_{y\to 0}(\lim_{x\to 0} f(x,y))$

$=\lim_{y\to0}(\lim_{x\to0} \frac{(y-x)}{(y+x)}×\frac{(1+x^2)}{(1+y^2)} )$

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

$=1$

Hence ,the repeated limits exist and unequal.

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

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

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

$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{x\to0} \frac{(mx-x)}{(mx+x)} ×\frac{(1+x^2)}{(1+m^2x^2)}$$

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

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

Hence the simultaneous limit does not exist.