Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$ I want to show that:
$$\lim_{(x,y)\to(0,0)} (1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} = 1$$
Direct substitution doesn't work, because of the denominator of $\frac{1}{x^2 + y^2}$.
Usually these problems are solved by bounding the following expression:
$$\left|(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} -1 \right| = \left| \frac{1 - (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}} }{ (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}}} \right|$$
I don't know how to continue.

Hint
Take the natural logarithm and use that $z-\frac{z^2}2\le \ln(1+z) \le z$.
Additionally we can switch to radial coordinates $x=r\cos\theta,\, y=r\sin\theta$ and let $r\to 0$.
