Determine if the following function is continuous on $ (0,0) $ Determine if the following function is continuous on $ (0,0) $
$$f(x,y)=\begin{cases}
 (1+x^2y^2)^{-1/(x^2+y^2)},& \text{ if } (x,y)\neq (0,0) \\ 
 1,& \text{ if } (x,y)=(0,0)
\end{cases}$$
Conjecturing that $ \lim _ {(x, y) \to (0,0)} f (x, y) = 1 $.
My attempt is,
$$\begin{align*}
0 &\leq | f (x) -L|\\
 &=|(1+x^2y^2)^{-1/(x^2+y^2)}-1|\\ &\leq |(1+x^2y^2)^{-1/(x^2+y^2)}|+1\end{align*}$$
Hence I do not know what else to do because in another limit like this but with the euler function, you can limit that using that the function is strictly increasing, but here you cannot do that because $ (1 + x ^ 2y ^ 2) $ is not strictly increasing , so I don't know how to limit that exponent with the function $ (1 + x ^ 2y ^ 2) $.
 A: $0 \leq \frac 1 {x^{2}+y^{2}} \ln (1+x^{2}y^{2}) \leq \frac 1 {x^{2}+y^{2}} x^{2}y^{2}\leq \frac 1  2|x||y| \to 0$. So $ \frac 1 {x^{2}+y^{2}} \ln (1+x^{2}y^{2}) \to 0$. Multiply by $-1$ and  take exponential.
A: Alternatively, you can use Bernoulli's inequality to obtain the desired result. We have:
$\left(1+x^2y^2\right)^{\frac{1}{x^2+y^2}}\ge 1+\dfrac{x^2y^2}{x^2+y^2}=\dfrac{x^2+y^2+x^2y^2}{x^2+y^2}\implies \left(1+x^2y^2\right)^{-\frac{1}{x^2+y^2}}\le\dfrac{x^2+y^2}{x^2+y^2+x^2y^2}= h(x,y)$. Also: $x^2+y^2 \ge x^2 \implies \dfrac{1}{x^2+y^2} \le \dfrac{1}{x^2}\implies -\dfrac{1}{x^2+y^2} \ge -\dfrac{1}{x^2}\implies \left(1+x^2y^2\right)^{-\frac{1}{x^2+y^2}}\ge \left(1+x^2y^2\right)^{-\frac{1}{x^2}} = \left(\left(1+x^2y^2\right)^{-\frac{1}{x^2y^2}}\right)^{y^2}=g(x,y)$. So: $g(x,y) \le f(x,y) \le h(x,y)$. But both $h,g \to 1$ when $(x,y) \to (0,0)$ because $0 \le \dfrac{x^2y^2}{x^2+y^2} \le x^2$, and thus $\dfrac{x^2y^2}{x^2+y^2} \to 0$. Thus $f(x,y) \to 1 = f(0,0)$, proving continuity of $f$ at $(0,0)$. Note that $g(x,y) \to e^{-0} = 1$ when $(x,y) \to (0,0)$.
