Let f ( x , y ) = ( x 2 + y 2 + 2 ) x y defined on R 2 and let g ( x , y ) be a degree 2 polynomial in two variables.Suppose lim ( x , y ) → ( 0 , 0 ) g ( x , y ) − f ( x , y ) x 2 + y 2 = 0 . Then, determine g ( x , y )

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Let   f  (  x  ,  y  )  =  (      x    2    +      y    2    +  2      )          x      y       defined on             R        2   and let   g  (  x  ,  y  ) be a degree 2 polynomial in two variables.Suppose             lim              (        x        ,        y        )        →        (        0        ,        0        )                                      g          (          x          ,          y          )          −          f          (          x          ,          y          )                                      x            2                    +                      y            2                                =    0  . Then, determine   g  (  x  ,  y  )

partyjnopp9wa · Accepted Answer

Recall that for       e    t    =  1  +  t  +  O  (      t    2    ) and   ln  ⁡  (  1  +  t  )  =  t  +  o  (  t  ). Therefore,for   (  x  ,  y  )  →  (  0  ,  0  )                    (                  x          2                +                  y          2                +        2                  )                      x            y                                              =        exp        ⁡        (        x        y        ln        ⁡        (                  x          2                +                  y          2                +        2        )        )                                          =        1        +        x        y        ln        ⁡        (                  x          2                +                  y          2                +        2        )        +        O        (                  x          2                          y          2                )                                          =        1        +        x        y                  (          ln          ⁡          (          2          )          +          ln          ⁡                      (            1            +                                                            x                  2                                +                                  y                  2                                            2                        )                    )                +        o        (                  x          2                +                  y          2                )                                          =        1        +        x        y                  (          ln          ⁡          (          2          )          +                                                    x                2                            +                              y                2                                      2                    +          o          (                      x            2                    +                      y            2                    )          )                +        o        (                  x          2                +                  y          2                )                                          =        1        +        ln        ⁡        (        2        )        x        y        +        o        (                  x          2                +                  y          2                )        .            Hence   g  (  x  ,  y  )  =  1  +  ln  ⁡  (  2  )  x  y. In particular       a    3    =  ln  ⁡  (  2  )

Let f ( x , y ) = ( x 2 </msup> + y 2 </msup> + 2

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