2

Can anyone show me the steps? The limit is $0$ but I am facing some difficulties in getting to that point! I know that $\ln(1+u) \leq u$ for $u>-1$.

$$\lim_{x,y\to 0}\frac{(x^3y+xy^3)\ln(1+x^2+y^4)}{x^4+6x^2y^2+y^4}$$

Integral
  • 6,554
Mustafa Motani
  • 123

1 Answers1

0

Hint: in polar coordinates, $\lim_{(x,y) \to (0,0)}$ becomes $\lim_{r \to 0^+}$ (where in general $\theta$ depends on $r$, if you do not approach the origin along a straight line). Now if $r \leq 1$, then using $\ln(1+x^2+y^4) \leq x^2+y^4$, you find that the numerator is less than $c_1 r^6$ while the denominator is at least $c_2 r^4$. So the quotient is at most $c_3 r^2$, where $c_1,c_2,c_3$ are constants.

Ian
  • 101,645
  • I am not sure I understood your solution. Are you at some point applying DeMoivre's theorem. btw thanks alot for your time...:) – Mustafa Motani Apr 15 '15 at 11:54
  • @MustafaMotani Replace $x$ with $r \cos(\theta(r))$ and $y$ with $r \sin(\theta(r))$. Use the log inequality that you stated, and send $r \to 0^+$. – Ian Apr 15 '15 at 12:00