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.

An alternative using polar coordinates:
Taking $x=r\cos \theta\;,y=r\sin\theta$
$$\lim_{r\to 0}(1+r^4\sin^2\theta\cos^2\theta)^{-1/r^2}=e^{\lim_{r\to 0}\frac{-1}{r^2}(r^4\sin^2\theta\cos^2\theta)}=e^{\lim_{r\to 0}(-r^2\sin^2\theta\cos^2\theta)}=e^0=1$$
