Answer to Question #339755 in Calculus for lily

Question

Show that the function f defined by

f(x,y) = {1, (x,y) = (0,0)

{(x² + y)/(x + y), (x,y) ≠ (0,0)

is not continuous by using two path test at (0,0).

Answer 1

If $x=0$ then $f(0, y)=\dfrac{0+y}{0+y}=1.$ Therefore

$f(x, y)\to 1\text{ as} \ (x,y)\to(0,0) \text{ along the }y-\text{axis}$

For all $x\not=0$

$f(x, x)=\dfrac{x^2+x}{x+x}=\dfrac{x+1}{2}$

$f(x, y)\to \dfrac{1}{2}\not=1\text{ as} \ (x,y)\to(0,0) \text{ along }y=x$

Since we have obtained different limits along different paths, limit

$\lim\limits_{(x,y)\to(0,0)}f(x, y)=\lim\limits_{(x,y)\to(0,0)}\dfrac{x^2+y}{x+y}$

does not exist.

The function

$f(x, y)= \begin{cases} 1, & (x,y)=(0,0) \\ \dfrac{x^2+y}{x+y}, &(x,y)\not=(0,0) \end{cases}$

is not continuous at $(0,0).$