Question

Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, explain why.​

Answer 1

(i) Yes. Simplify
`f(x,y)`.

`\displaystyle (x^2 - x^2y^2 + y^2)/(x^2 + y^2) = 1 - (x^2y^2)/(x^2 + y^2)`

Now compute the limit by converting to polar coordinates.

`\displaystyle \lim_((x,y)\to(0,0)) (x^2y^2)/(x^2+y^2) = \lim_(r\to0) (r^4 \cos^2(\theta) \sin^2(\theta))/(r^2) = 0`

This tells us

`\displaystyle \lim_((x,y)\to(0,0)) f(x,y) = 1`

so we can define
`f(0,0)=1` to make the function continuous at the origin.

Alternatively, we have

`(x^2y^2)/(x^2+y^2) \le (x^4 + 2x^2y^2 + y^4)/(x^2 + y^2) = ((x^2+y^2)^2)/(x^2+y^2) = x^2 + y^2`

and

`(x^2y^2)/(x^2+y^2) \ge 0 \ge -x^2 - y^2`

Now,

`\displaystyle \lim_((x,y)\to(0,0)) -(x^2+y^2) = 0`

`\displaystyle \lim_((x,y)\to(0,0)) (x^2+y^2) = 0`

so by the squeeze theorem,

`\displaystyle 0 \le \lim_((x,y)\to(0,0)) (x^2y^2)/(x^2+y^2) \le 0 \implies \lim_((x,y)\to(0,0)) (x^2y^2)/(x^2+y^2) = 0`

and
`f(x,y)` approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

`\displaystyle (x^2 + y^3)/(xy) = \frac xy + \frac{y^2}x`

`f(0,y)` and
`f(x,0)` are undefined, so there is no way to make
`f(x,y)` continuous at (0, 0).

(iii) No. Similarly,

`\frac{x^2 + y}y = \frac{x^2}y + 1`

is undefined when
`y=0`.