Find the limit : $\lim_{(x,y) \to (+\infty,+\infty)}(x^2+y^2)e^{-(x+y)}$ Evaluate:
$$\lim_{(x,y) \to (+\infty,+\infty)}(x^2+y^2)e^{-(x+y)}$$

We have,
$$\begin{align}\lim_{(x,y)\to (\infty,\infty)}\frac{x^2+y^2}{e^{x+y}}&=\lim_{(x,y)\to (\infty,\infty)}\frac{(x+y)^2-2xy}{e^{x+y}}\\
&=\lim_{(x,y)\to (\infty,\infty)} \frac{(x+y)^2}{e^{x+y}}-\lim_{(x,y)\to (\infty,\infty)} \frac{2xy}{e^{x+y}}\\
&=-2\lim_{(x,y)\to (\infty,\infty)} \frac{xy}{e^{x+y}}\\
&=-2 \lim_{x\to \infty}\frac{x}{e^x}\times \lim_{y\to \infty}\frac{y}{e^y}\\
&=0.\end{align}$$

I did not work with the limits with two variables.  That's why, I'm not sure I made a rigorous proof.
My question is:

*

*Is this method mathematically valid?  Are there any non-rigorous steps in my method?

Thank you.
 A: Using $x^2+y^2 \le (x+y)^2$ for $x,y>0$ we have $(x^2+y^2)e^{-(x+y)} \le (x+y)^2 e^{-(x+y)}$
So $$\lim_{(x,y)\to(\infty,\infty)} (x^2+y^2)e^{-(x+y)} \le \lim_{(x,y)\to(\infty,\infty)}  (x+y)^2 e^{-(x+y)} = \lim_{u\to\infty} u^2e^{-u} = 0$$
A: Your solution seem fine. Your breaking up of the limits is perfectly valid.
I would have approached this by noting that, by using L'Hôspital twice, we get
$$
\lim_{u\to\infty}\frac{u^2}{e^u}=0\tag1
$$
Then, because $(x,y)$ is eventually in the first quadrant, $(x,y)=r(\cos(\theta),\sin(\theta))$, where $0\le\theta\le\frac\pi2$.
$$
\begin{align}
\frac{x^2+y^2}{(x+y)^2}
&=\frac1{(\cos(\theta)+\sin(\theta))^2}\tag{2a}\\
&=\frac1{2\sin^2\left(\theta+\frac\pi4\right)}\tag{2b}\\[3pt]
&\le1\tag{2c}
\end{align}
$$
Let $u=x+y$, then as $(x,y)\to(\infty,\infty)$, $u\to\infty$.
$$
\begin{align}
\lim_{(x,y)\to(\infty,\infty)}\left(x^2+y^2\right)e^{-x-y}
&\le\limsup_{(x,y)\to(\infty,\infty)}\frac{x^2+y^2}{(x+y)^2}\lim_{(x,y)\to(\infty,\infty)}(x+y)^2e^{-x-y}\tag{3a}\\
&\le\sup_{0\le\theta\le\pi/2}\frac1{(\sin(\theta)+\cos(\theta))^2}\lim_{u\to\infty}u^2e^{-u}\tag{3b}\\[3pt]
&=1\cdot0\tag{3c}
\end{align}
$$

Graphical Solution Verification
The first term on the right side of $\text{(3a)}$ is

The second term on the right side of $\text{(3a)}$ is

Thus, as $(x,y)\to(\infty,\infty)$, we have
$$
\overbrace{\frac{x^2+y^2}{(x+y)^2}}^{\le1}\overbrace{(x+y)^2e^{-x-y}\vphantom{\frac{x^2+y^2}{(x+y)^2}}}^{\to0}\to0
$$
A: I am assuming that you are talking about the limit at $(+\infty, +\infty)$.
If I understood your intent correctly, I think your idea could be made rigorous by using sequences : take any $(x_n, y_n)_n$ that tends to $(+\infty, +\infty)$, and check that the sequence $(x_n^2 + y_n^2)e^{-(x_n+y_n)}$ tends to $0$.
More precisely : you can define the quantities $u_n$ and $v_n$ that correspond to your original idea. Then use operations on limits to adapt your previous informal ideas.
For instance, the final steps in your proof would become : $\frac{v_n}{e^{u_n}}= \frac{x_n y_n}{e^{x_n} e^{y_n}} \rightarrow 0$ because $\frac{x_n}{e^{x_n}} \rightarrow 0$ and  $\frac{y_n}{e^{y_n}} \rightarrow 0$.
The general fact I am using here : If $f$ is a function (possibly with several variables), if $l, a$ are numbers (or $\pm\infty$), then $f(z)$ tends to $l$ when $z$ tends to $a$ if and only if, for all sequences $(z_n)$ that tend to $l$ such that $f(z_n)$ is well-defined, $f(z_n)$ tends to $l$ as $n$ tends to infinity.
This is true even in more than one variable, as long as there are finitely many variables and you are dealing with real or complex numbers (I am not getting into the general topological definition, unless you are really interested).
A: If we let $x = r\cos\theta,y = r\sin\theta$, then $\theta\in (0,\pi/2),r\to\infty$ as $(x,y)\to(+\infty,+\infty)$. Now
$$\lim_{(x,y)\to(+\infty,+\infty)}(x^2+y^2)e^{-(x+y)} = \lim_{(r,\theta)\to(\infty,\theta_0)}r^2e^{-r(\cos\theta+\sin\theta)}\leq \lim_{(r,\theta)\to(\infty,\theta_0)} r^2e^{-r} =0$$
where $\theta_0\in (0,\pi/2)$. I wonder such expression is valid. The notion $\theta_0$ means that it may not converge to particular value but should be in $(0,\pi/2)$.
