<s>[INST] 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.
 [/INST] 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
$$
\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left( {1 + x^2 y^2 } \right)^{ - \frac{1}{{x^2  + y^2 }}}  = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} e^{ - \frac{1}{{x^2  + y^2 }}\log \left( {1 + x^2 y^2 } \right)}  = e^0  = 1
$$
</s>