# Does $\lim\limits_{y\to 0}\space\lim\limits_{x\to\infty}xy=0,1,\text{or}\space\infty?$ and a more specific question

$$\lim\limits_{y\to 0}\space\lim\limits_{x\to\infty}xy=0,1,\text{or}\space\infty?$$

This problem occurred as I came to the double integral $$\int_0^1\int_1^\infty f(x,y)dxdy$$ where $f(x,y)$ includes an $xy$ in it's definition. After taking both indefinite integrals, the $xy$ remains in the definition. In the case of $f(x,y)=xy$ it appears that the double limit is zero, but is this always the case? And how do we prove it?

For context, my two integrals that cause this problem are $$\lim\limits_{x\to\infty}\int_0^1\frac{y^4\left(y^2-3\right)}{\left(y^2+1\right)^3}{_2F_1}\left(1,\frac{a+1}{2};\frac{a+3}{2};-x^2y^2\right)dy$$ and $$\lim\limits_{x\to\infty}\int_0^1\frac{y^4\left(3y^2-1\right)}{\left(y^2+1\right)^3}{_2F_1}\left(1,\frac{a+2}{2};\frac{a+4}{2};-x^2y^2\right)dy$$ In the step for both of these expressions, the first integral has been done and the limit shows that we are considering the first integral's evaluation at $\infty$ with respect to $x$. And $_2F_1(a,b;c;z)$ is the Hypergeometric function.

How should I do this? Should I make it easier and limit $x\to\infty$ first? Otherwise, I have no idea how to integrate this keeping it in terms of $a$ and $x$.

• Well, to answer the title question, $\lim_{x\to \infty} xy$ is something like $\operatorname{sgn}(y)\infty$ ($0$ if $y = 0$), so $\lim_{y\to 0}\lim_{x\to \infty} xy$ won't exist even in the context of infinite limits. – Michael Lee Aug 22 '17 at 15:49

$\lim\limits_{y\to0}\lim\limits_{x\to\infty}xy$ is the limit for $y\to\infty$ of a function $f:y\mapsto\lim\limits_{x\to\infty}xy$. This function is only defined for $y=0$ because otherwise the function $g_y:x\mapsto xy$ doesn't have an asymptote. So the domain of $f$ consists only of the point $0$. Therefore the limit is not defined because the point at which you take the limit needs to be a limit point of the $f$'s domain, which $0$ is not. If you, however, insist on getting an answer apart from undefined, you can drop the limit point requirement and take the limit to be $f(0)=0$.