I am trying to solve an exercise to show that this function f ( x , y ) = x 2 + y 2 x 2 + y 2 + 1 − 1

Question

I am trying to solve an exercise to show that this function  f  (  x  ,  y  )  =                    x        2            +              y        2                                      x          2                +                  y          2                +        1            −      1

Reagan Madden · Accepted Answer

You have                                                        x              2                        +                          y              2                                                                          x                2                            +                              y                2                            +              1                        −            1                                              =                              (                          x              2                        +                          y              2                        )                          (                                                x                  2                                +                                  y                  2                                +                1                            +              1              )                                                          (                                                x                  2                                +                                  y                  2                                +                1                            −              1              )                                      (                                                x                  2                                +                                  y                  2                                +                1                            +              1              )                                                                        =                              x            2                    +                      y            2                    +          1                +        1            and therefore      |                            x          2                +                  y          2                                                  x            2                    +                      y            2                    +          1                −        1              −    2    |    =      |                  x        2            +              y        2            +      1        −    1    |    .and therefore      |                            x          2                +                  y          2                                                  x            2                    +                      y            2                    +          1                −        1              −    2    |    =      |                  x        2            +              y        2            +      1        −    1    |    .But      |                  x        2            +              y        2            +      1        −    1    |    ⩽            x      2        +          y      2        .In fact, this inequality is equivalent to            x      2        +          y      2        +  2  −  2            x      2        +          y      2        +    1    ⩽            x      2        +          y      2      and it is clear that we always have   2  ⩽  2            x      2        +          y      2        +    1  Putting all together, we have      |                            x          2                +                  y          2                                                  x            2                    +                      y            2                    +          1                −        1              −    2    |    ⩽            x      2        +          y      2      and so, in order to prove that      lim          (      x      ,      y      )      →      (      0      ,      0      )                          x        2            +              y        2                                      x          2                +                  y          2                +        1            −      1        =  2  ,for each   ε  &amp;gt;  0, we only have to take   δ  =  ε

Leland Morrow · Accepted Answer

Let       r    2    =      x    2    +      y    2        |              r      2                                r          2                +        1            −      1        −  2      |    &amp;lt;  ϵ  2  −  ϵ  &amp;lt;            r      2        +    1    +  1  &amp;lt;  2  +  ϵ by rationalizing the denominator, cancelling terms and pushing 2 to the outsides of the compound inequality.            r      2        +    1    ≤  1  +            r      2        2      ⟹    −  ϵ  &amp;lt;      r    2        /    2  &amp;lt;  ϵ by Taylor&#039;s Theorem, then cancelling, then swapping x and y back in.      x    2    +      y    2    &amp;lt;  2  ϵ. So let   δ  =  ϵ for a simpler, tighter bound.

I am trying to solve an exercise to show that this function f ( x , y ) =

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