Calculating $\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$ Please help me calculate the:
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$$
My attempt is: First method: for $t>0$ we have $e^t\geq \frac{t^3}{3}$ now i have:
$$(x^2+y^2)e^{-(x+y)}\leq\frac{3(x^2+y^2)}{(x+y)}\rightarrow 0, (x,y)\rightarrow(\infty,\infty)$$
but i dint know how to continue
second method:the polar coord
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}=\lim_{r\rightarrow \infty}(r^2)e^{-r(cos\theta+sin\theta)}=\lim_{r\rightarrow \infty}\frac{r^2}{e^{r(cos\theta+sin\theta)}}=\frac{1}{cos\theta+\sin\theta}\lim_{r\rightarrow \infty}\frac{2r}{e^r}=\frac{2}{cos\theta+\sin\theta}\lim_{r\rightarrow \infty}\frac{1}{e^r}=\frac{2}{cos\theta+\sin\theta}\cdot0=0$$ i use the L'Hopital rule,but i dint know its correct

Using $ e^t > \frac{t^4}{4!}$, we have for $x,y >0$
$$e^{x+y} > \frac{(x+y)^4}{4!} = \frac{(x^2+y^2+2xy)^2}{4!} > \frac{(x^2+y^2)^2}{4!} $$
$$(x^2+y^2)e^{-(x+y)} < 4! \frac{x^2+y^2}{(x^2+y^2 )^2}=\frac{4!}{(x^2+y^2)} \to 0$$
