Question

Find the value c that makes f(x, y) a continuous function.​

Answer 1

To ensure continuity at the origin, we need to first show that the limit

`\displaystyle \lim_((x,y)\to(0,0)) (x^2+y^2)/(√(x^2+y^2+1) - 1)`

exists.

This is easy to do - simplifying
`f(x,y)`, we get

`(x^2 + y^2)/(√(x^2 + y^2 + 1) - 1) = ((x^2+y^2) \left(√(x^2 + y^2 + 1) + 1\right))/(\left(√(x^2+y^2+1)\right)^2-1^2) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~= ((x^2 + y^2) \left(√(x^2+y^2+1)+1\right))/(x^2+y^2) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~= √(x^2 + y^2 + 1) + 1`

when
`x^2+y^2\\eq0`. The simplified form is continuous for all
`(x,y)\in\Bbb R^2`, so

`\displaystyle \lim_((x,y)\to(0,0)) f(x,y) = \lim_((x,y)\to(0,0)) \left(√(x^2+y^2+1) + 1\right) = 2`

and we should choose
`\boxed{c=2}`.