-2

It is part of my HW, I solved it and got that the following limit is 2, But Wolfram says that the limit does not exists \ path dependent. After few hours of trying to find a path to dismiss the limit I came up with nothing..

Would appreciate any help (sorry in advance for not writing it nicely xD )

$\lim_ {(x,y) \rightarrow(0,0)} \frac{\ln(1+2x^2+4y^2)}{\arctan(x^2+2y^2)}$

azimut
  • 22,696
donKipod
  • 11
  • 1
  • My answer for http://math.stackexchange.com/questions/572125/show-discontinuity-of-fracxyx2y2/572148#572148 might be helpful to some extent.. –  Nov 22 '13 at 10:54
  • 1
    What paths did you use? – egreg Nov 22 '13 at 11:50

1 Answers1

2

Consider writing your expression as Log[1+2z]/ArcTan[z]."z" being small, replace both numerator and denominator by the first order Taylor expansion. You then get a limit equal to 2. If you draw a 3D plot of your function fo x and y (say between - 1 and + 1), it is a very nice surface (cowboy hat ?) showing a maximum value at [0,0].

Claude Leibovici
  • 260,315