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 answer without polar coordinates, using bounds as OP was interested in. The Bernoulli inequality yields $(1+x^2y^2)^{(x^2+y^2)^{-1}}\ge1+\frac{x^2y^2}{x^2+y^2}$ so we can straightaway bound: $$\left|\frac{1-(1+x^2y^2)^{(x^2+y^2)^{-1}}}{(1+x^2y^2)^{(x^2+y^2)^{-1}}}\right|\le\frac{|1-(1+x^2y^2)^{(x^2+y^2)^{-1}}|}{1+\frac{x^2y^2}{x^2+y^2}}$$
Similarly using a variant of the Bernoulli inequality ($(1+x)^r\le e^{xr}$ for $r\gt0$, $x\gt0$) and choosing $(x,y)$ in a neighbourhood of the origin so that the ratio $\frac{x^2y^2}{x^2+y^2}\lt1$: $$0\le\frac{(1+x^2y^2)^{(x^2+y^2)^{-1}}-1}{1+\frac{x^2y^2}{x^2+y^2}}\le\frac{e^{\frac{x^2y^2}{x^2+y^2}}-1}{1+\frac{x^2y^2}{x^2+y^2}}\le\frac{\frac{x^2y^2}{x^2+y^2}(e-1)}{1+\frac{x^2y^2}{x^2+y^2}}$$So we are now interested in showing that: $$\frac{x^2y^2}{x^2+y^2+x^2y^2}\overset{(x,y)\to0}{\longrightarrow}0$$
On any path of approach with $|x|\le|y|$ we have: $$\frac{x^2y^2}{x^2+y^2+x^2y^2}\le\frac{y^2}{1+1+y^2}\to0$$Symmetrically we get the same when approaching $|y|\le|x|$. The "sequence of approach" criterion of limits in this space concludes.
