<s>[INST] 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
 [/INST] 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
$$(x^2+y^2)e^{-(x+y)}\le{3(x^2+y^2)\over(x+y)^3}={3x^2\over(x+y)^3}+{3y^2\over(x+y)^3}\le{3x^2\over x^3}+{3y^2\over y^3}={3\over x}+{3\over y}\to0+0=0$$
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