Show function is continuous at $(0,0)$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function with
$$\space f(x, y) = \begin{equation}
\begin{cases}
\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}, & (x, y) \neq (0, 0)\\
0, & (x, y) = 0 
\end{cases}\end{equation}$$
Show that f is continious.
I've already shown that f is continous for $(x, y) \neq (0,0)$ but I am having trouble with finding a proof for $(0, 0)$ using epsilon-delta or limits of sequences. Can anyone help? 
This is how far I got with simplifying for epsilon delta
$$
\Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}\Bigg{|} \\ 
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{y^4*(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{y^2\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{\log(1+x^2y^2)}{\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\leq \big{|}\log(1+x^2y^2)\big{|}
$$
But now I am clueless on how to connect
$$
\big{|}log(1+x^2y^2)\big{|} < \epsilon \\
$$
and
$$
\sqrt{x^2+y^2} < \delta
$$
 A: Ιf you use polar coordinates you will have that $$f(r,t)=\frac{1}{\sqrt{2}}\log{(1+r^4\cos^2{t}\sin^2{t}})\to^{r \to 0}0$$
So the limit as $(x,y) \to (0,0)$ is zero.
A: Hint: $\log (1+u)\le u$ for $u\ge 0.$
A: To conclude your proof we can use that by AM-GM
$$x^2y^2 \le \left(\frac{x^2+y^2}2\right)^2$$
$$\big{|}\log(1+x^2y^2)\big{|} \le x^2y^2 \le \left(\frac{x^2+y^2}2\right)^2=\frac{\left(\sqrt{x^2+y^2}\right)^4}{4}<\epsilon $$
as $\sqrt{x^2+y^2} < \sqrt[4]{4\epsilon}$.
As an alternative, we have that
$$\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}=\dfrac{\log(1+x^2y^2)}{x^2y^2}\dfrac{x^2y^4}{\sqrt{x^4+y^4}}\to 0$$
indeed by standard limits $\dfrac{\log(1+x^2y^2)}{x^2y^2} \to 1$ and
$$\dfrac{x^2y^4}{\sqrt{x^4+y^4}}=y^4\dfrac{\sqrt{x^4}}{\sqrt{x^4+y^4}}\le y^4 \to 0$$
A: Let $|x|,|y| <1$.
$\log (1+x^2y^2) \le x^2y^2 \le x^4 +y^4 \le (x^2+y^2);$
$\epsilon >0$ be given.
Choose $\delta =\epsilon^{1/2}$.
Used 
AM-GM: $x^4+y^4 \ge 2x^2y^2$; and
$\log (1+z) \le z$ for $z \ge -1$.
A: in the numerator  $0\le y^2\log(1+x^2y^2)\le x^2y^4$
In the denominator $\sqrt{x^4+y^4} \ge  \max (x^2,y^2)\ge 0$
