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.
 A: 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 $r\to 0$.
A: 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.
A: An alternative using polar coordinates:
Taking $x=r\cos \theta\;,y=r\sin\theta$
$$\lim_{r\to 0}(1+r^4\sin^2\theta\cos^2\theta)^{-1/r^2}=e^{\lim_{r\to 0}\frac{-1}{r^2}(r^4\sin^2\theta\cos^2\theta)}=e^{\lim_{r\to 0}(-r^2\sin^2\theta\cos^2\theta)}=e^0=1$$
A: You can also observe that
$$
\begin{array}{l}
 \left( {1 + x^2 y^2 } \right)^{ - \frac{1}{{x^2  + y^2 }}}  = e^{ - \frac{1}{{x^2  + y^2 }}\log \left( {1 + x^2 y^2 } \right)}  =  \\ 
  \\ 
  = e^{ - \frac{{x^2 y^2 }}{{x^2  + y^2 }}\frac{{\log \left( {1 + x^2 y^2 } \right)}}{{x^2 y^2 }}}  \\ 
 \end{array}
$$
Since
$$
0 \le \frac{{x^2 y^2 }}{{x^2  + y^2 }} = x^2 \frac{{y^2 }}{{x^2  + y^2 }} \le x^2 ,\,\,\,\,\forall \left( {x,y} \right) \in \mathbb R^2  \setminus \left( {0,0} \right)
$$
you have
$$
\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} x^2 \frac{{y^2 }}{{x^2  + y^2 }} = 0
$$
Moreover
$$
\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\log \left( {1 + x^2 y^2 } \right)}}{{x^2 y^2 }} = \mathop {\lim }\limits_{t \to 0} \frac{{\log \left( {1 + t} \right)}}{t} = 1
$$
Thus
$$
\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{x^2 y^2 }}{{x^2  + y^2 }}\frac{{\log \left( {1 + x^2 y^2 } \right)}}{{x^2 y^2 }} = 0
$$
and
$$
\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left( {1 + x^2 y^2 } \right)^{ - \frac{1}{{x^2  + y^2 }}}  = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} e^{ - \frac{1}{{x^2  + y^2 }}\log \left( {1 + x^2 y^2 } \right)}  = e^0  = 1
$$
