Limit of a 2-variable function Given problem:
$$
\lim_{x\to0,y\to0}(1+x^2y^2)^{-1/(x^2+y^2)} 
.$$
I tried to do it with assigning y to $y = kx$, but that didn't help me at all. Also one point, I can't use L'Hospital's rule.
 A: Using Polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$, the problem boils down to $$ \lim_{r\to0}\space(1 + r^4\cos^2\theta\sin^2\theta)^{\frac{-1}{r^2}}$$
Now prove that independent of the value of $\theta$, the limit exists and solve for it then.
EDIT:
Now let $$L = \lim_{r\to0}\space(1 + r^4\cos^2\theta\sin^2\theta)^{\frac{-1}{r^2}}$$
Taking $\log$ on both sides,
$$\log L = \lim_{r\to0}\space -\dfrac{\log(1 + r^4\cos^2\theta\sin^2\theta)}{r^2}$$As $\log (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$,
$$\log L = \lim_{r\to0}\space - \dfrac{1}{r^2}\left (r^4\cos^2\theta\sin^2\theta - \dfrac{\left (r^4\cos^2\theta\sin^2\theta\right )^2}{2}) + \cdots \right ) = 0$$
$$\implies \log L =0 \implies L = 1$$
A: Note that we can write
$$(1+x^2y^2)^{-1/(x^2+y^2)}=e^{-\frac{\log(1+x^2y^2)}{x^2+y^2}}$$

Recall that the logarithm satisfies the inequalities (SEE THIS ANSWER)
$$\frac{x-1}{x}\le\log(x)\le x-1 \tag 1$$

Applying the right-hand side inequality in $(1)$ along with the AM-GM inequality leads to the estimate
$$\begin{align}
\frac{\log(1+x^2y^2)}{x^2+y^2}&\le \frac{x^2y^2}{2|x||y|}\\\\
&=\frac{|xy|}2 \to 0
\end{align}$$

Finally, using continuity of the exponential function, we obtain the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{(x,y)\to(0,o0)}(1+x^2y^2)^{-1/(x^2+y^2)}=1}$$
A: I assume here that you mean $\lim_{(x,y) \to (0,0)}$ then 
$$
\begin{align}
\lim_{(x,y) \to (0,0)} (1+x^2y^2)^{-\frac{1}{x^2+y^2}} = 
 \lim_{(x,y) \to (0,0)} \exp\left( \ln \left( (1+x^2y^2)^{-\frac{1}{x^2 + y^2}} \right) \right) &= \\ \\\lim_{(x,y) \to (0,0)} \exp\left( -\frac{\ln  (1+x^2y^2)}{x^2 + y^2}  \right) &= \\ \exp\left(  -\lim_{(x,y) \to (0,0)} \frac{\ln  (1+x^2y^2)}{x^2 + y^2}  \right) \tag{1}
\end{align}
$$
Assuming the limit exists and by continuity of the exponential.
Now if we look at the function which is clearly non-negative and using the fact that $\ln(1+x) \leq x$ we have 
$$
 0 \leq \frac{\ln  (1+x^2y^2)}{x^2 + y^2} \leq \frac{x^2y^2}{x^2+y^2} \\
 \lim_{(x,y) \to (0,0)} 0 \leq  \lim_{(x,y) \to (0,0)}\frac{\ln  (1+x^2y^2)}{x^2 + y^2} \leq \lim_{(x,y) \to (0,0)} \frac{x^2y^2}{x^2+y^2} = \\
 \lim_{r\to 0}\frac{r^4 \cos(t)^2\sin(t)^2}{r^2} = 0
$$
Hence 
$$
\lim_{(x,y) \to (0,0)}\frac{\ln  (1+x^2y^2)}{x^2 + y^2}  = 0 \tag{2}
$$
$(1)$ and $(2)$ allow you to conclude.
A: The expression equals
$$\tag 1\left ((1+x^2y^2)^{1/(x^2y^2)}\right )^{-x^2y^2/(x^2+y^2)}.$$
Because $(1+u)^{1/u} \to e$ as $u\to 0,$ the expression inside the parentheses $\to e.$ Since
$$0\le x^2y^2/(x^2+y^2) \le x^2[y^2/(x^2+y^2)] \le x^2,$$
the outside exponent in $(1)$ goes to $0.$ The desired limit is therefore $e^0=1.$
(Note: There's a small problem above. What if $x^2y^2=0,$ which would render $1/(x^2y^2)$ meaningless? That's really no problem because the expression of interest actually equals $1$ when $x^2y^2=0.$)
