Evaluating $\lim_{(x,y)\to(0,0)}(x^2+y^2)^{x^2y^2}$

Question

Been trying to solve this multivariable limit but I’m stuck.

$$\lim_{(x,y)\to(0,0)}(x^2+y^2)^{x^2y^2}$$

I know that, in one dimension, one can use $$x^y=e^{y\log x}$$ but unfortunately, I haven’t been told how to proceed in higher dimension.

Answer 1

HINT

Note that by polar coordinates

• $x=r\cos \theta$
• $y=r\sin \theta$

$$\large(x^2+y^2)^{x^2y^2}=r^{2r^4\cos^2\theta\sin^2\theta}=e^{(2r^3\cos^2\theta\sin^2\theta)(r\log r)}$$

then evaluate as $r\to 0$

• $2r^3\cos^2\theta\sin^2\theta$
• $r\log r$

Answer 2

Apply $\ln $ to the expression to get

$$\tag 1 x^2y^2\ln (x^2+y^2) = \frac{x^2y^2}{x^2+y^2}(x^2+y^2)\ln (x^2+y^2).$$

If $|y|\le 1,$ then the fraction on the right is bounded above by $x^2/(x^2+y^2) \le 1.$ For these $y$ the absolute value of $(1)$ is bounded above by $(x^2+y^2)|\ln (x^2+y^2)|.$ Now use $u|\ln u| \to 0$ as $u\to 0^+$ to see that $(1)\to 0.$ Exponentiating back gives $1$ for the desired limit.