Check whether the given function is continuous or not at the point (0, 0)

Check whether the given function is continuous or not at the point (0, 0).

Solution:

We know that A function f of two variables is called continuous at the point `(x_0, y_­­0)` if

`f(x_0 , y_­­0)`  is defined.

` lim_{(x,y) \to (x_0,y_0)} f(x,y)`exists.

` lim_{(x,y) \to (x_0,y_0)} f(x,y) = f(x_0,y_0)`

To determine,  if f is continuous at (0,0), we need to compare ` lim_{(x,y) \to (x_0,y_0)} f(x,y) ` to  `f(x_0,y_0)`

According to the definition of the function f, we find that

`f(0,0) = 1`

we now consider the limit ` lim_{(x,y) \to (x_0,y_0)} f(x,y)` of the function

`f(x,y) = \frac{x^2 + y^2 - x^2y^2}{x^2 + y^2}`

Let (x,y) approaches to the line x, where y=0

`f(x,y) = \frac{x^2 + 0^2 - x^20^2}{x^2 + 0^2}`

`f(x,y) = \frac{x^2 + 0 - 0}{x^2 + 0}`

`f(x,y) = \frac{x^2}{x^2}`

`f(x,y) = 1                 -  -  -  - (1)`

Let (x,y) approaches to the line y, where x=0

`f(x,y) = \frac{0^2 + y^2 - 0^2y^2}{0^2 + y^2}`

`f(x,y) = \frac{0 + y^2 - 0}{0 + y^2}`

`f(x,y) = \frac{y^2}{y^2}`

`f(x,y) = 1           -  -  -  -  - ( 2 )`

Let (x,y) approaches to the line, where y=x

`f(x,y) = \frac{x^2 + x^2 - x^2x^2}{x^2 + x^2}`

`f(x,y) = \frac{2x^2 - x^4}{2x^2}`

`f(x,y) = \frac{2x^2}{2x^2} - \frac{x^4}{2x^2}`

`f(x,y) = 1 - \frac{x^2}{2}`

By putting `f(x_0 , y_­­0) = (0,0)`

`f(x,y) = 1 - \frac{0^2}{2}`

`f(x,y) = 1           -  -  -  -  - ( 3 )`

From Equation (1),(2) and (3),

It is proved that the limit of the Above function exists and is equal to 1.

`f(x_0 , y_­­0) = 1`

`lim_{(x,y) \to (x_0,y_0} \right)  f(x,y) = 1`

`lim_{(x,y) \to (x_0,y_0} \right)  f(x,y) =f(x_0 , y_­­0) = 1`

Hence,

It is proved that the given function 

is continuous or not at the point (0,0)