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
 A: Your method of showing that $r^2\exp(-r[\cos\theta+\sin\theta])$ is also correct. The only thing you need to consider to the finish line, is that the angle $\theta$ must fall within $[0,\pi/2]$, causing the term $\cos\theta+\sin\theta$ to be at least $\sqrt 2/2$. Then you can conclude the final limit.
A different approach can be taken by noting that for $x,y>0$
$$
0<(x^2+y^2)e^{-x-y}=x^2e^{-x-y}+y^2e^{-x-y}<x^2e^{-x}+y^2e^{-y}
$$
and the conclusion comes very easily.
A: Using $ e^t > \frac{t^4}{4!}$, we have for $x,y >0$
$$e^{x+y} > \frac{(x+y)^4}{4!} = \frac{(x^2+y^2+2xy)^2}{4!} > \frac{(x^2+y^2)^2}{4!} $$
$$(x^2+y^2)e^{-(x+y)} < 4! \frac{x^2+y^2}{(x^2+y^2 )^2}=\frac{4!}{(x^2+y^2)} \to 0$$
A: Aside from what I take to be a typo (the denominator under the $3(x^2+y^2)$ should be $(x+y)^3$, not just $(x+y)$), your first method looks fine. It could be elaborated to
$$(x^2+y^2)e^{-(x+y)}\le{3(x^2+y^2)\over(x+y)^3}={3x^2\over(x+y)^3}+{3y^2\over(x+y)^3}\le{3x^2\over x^3}+{3y^2\over y^3}={3\over x}+{3\over y}\to0+0=0$$
