Calculate $\lim_{(x,y)\to (0,0)}\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1}$ How can I calculate the limit of the following?
$$ \lim_{(x,y)\to (0,0)}\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1} $$
I've tried using the sandwich rule, but after more than 2 hours, with no success.
Would my appreciate help.
Thank you!
 A: We have that
$$\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1}=\frac{\arctan(x^2+y^4) }{ x^2+y^4}\cdot\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}$$
with
$$\frac{\arctan(x^2+y^4) }{ x^2+y^4}\to 1$$
and using polar coordinates we obtain
$$\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}=r^2\frac{\cos^2\theta+r^2\sin^4\theta }{ \sqrt{r^2+1} - 1}=r^2\frac{\cos^2\theta+r^2\sin^4\theta }{ \sqrt{r^2+1} - 1}\frac{\sqrt{r^2+1} + 1 }{ \sqrt{r^2+1} + 1}=$$
$$= r^2\frac{\cos^2\theta+r^2\sin^4\theta }{r^2}(\sqrt{r^2+1} + 1)=(\cos^2\theta+r^2\sin^4\theta)(\sqrt{r^2+1} + 1)\to 2\cos^2 \theta$$
which is dependent by $\theta$.

To see that without polar coordinates let consider

*

*$x=y=t \to 0$ then

$$\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}=\frac{t^2+t^4 }{ \sqrt{2t^2+1} - 1}=\frac{t^2+t^4 }{ \sqrt{2t^2+1} - 1}\cdot \frac{\sqrt{2t^2+1} + 1}{ \sqrt{2t^2+1} + 1}=$$
$$=\frac{t^2(1+t^2)(\sqrt{2t^2+1} + 1)}{ 2t^2}=\frac12(1+t^2)(\sqrt{2t^2+1}+1)\to 1$$

*

*$x=t \to 0$ and $y=0$ then

$$\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}=\frac{t^2 }{ \sqrt{t^2+1} - 1}=\frac{t^2 }{ \sqrt{t^2+1} - 1}\cdot \frac{\sqrt{t^2+1} + 1}{ \sqrt{t^2+1} + 1}=$$
$$=\frac{t^2(\sqrt{t^2+1} + 1)}{ t^2}=\sqrt{t^2+1}+1\to 2$$
